Employ the following methods to find the maximum of the function from Prob. 13.7: (a) Golden-section search ( x l = − 2 , x u = 1 , ε s = 1 % ) . (b) Parabolic interpolation ( x 0 = − 2 , x 1 = − 1 , x 2 = 1 , iterations = 4 ) . Select new points sequentially as in the secant method. (c) Newton's method ( x 0 = − 1 , ε s = 1 % )

Question

Want to see more full solutions like this?

Answer 1

Textbook Question

Chapter 13, Problem 8P

Employ the following methods to find the maximum of the function from Prob. 13.7:

(a) Golden-section search ( x l = − 2 , x u = 1 , ε s = 1 % ) .

(b) Parabolic interpolation ( x 0 = − 2 , x 1 = − 1 , x 2 = 1 , iterations = 4 ) . Select new points sequentially as in the secant method.

(c) Newton's method ( x 0 = − 1 , ε s = 1 % )

(a)

Expert Solution

To determine

To calculate: The maximum of the function f(x)=−x4−2x3−8x2−5x with the help of Golden-section search. Take initial guesses as xl=−2,xu=1, andan acceptable error as et=1%.

Answer to Problem 8P

Solution:

The maximum of the function f(x)=−x4−2x3−8x2−5x is, x=−0.3468 after 14 iterations.

Explanation of Solution

Given information:

The function f(x)=−x4−2x3−8x2−5x. Initial guesses xl=−2,xu=1 and value of acceptable error is et=1%.

Formula used:

The golden-search method with two initial guesses, xl and xu that encloses one local extreme of function f(x). Two interior points x1 and x2 with the golden ratio d is given by,

d=5−12( xu−xl)x1=xl+dx2=xu−d

Evaluated the function at the above two interior points. Two results can occur,

If f(x1)>f(x2), then x2 becomes the new xl for the next iteration.

If f(x2)>f(x1), then x1 becomes the new xu for the next iteration.

Calculation:

Consider function f(x), in standard form as

f(x)=−x4−2x3−8x2−5x

With xl=−2 , xu=1 and εs=1%

Iteration 1: First golden ratio is used to create two interior points as,

d=5−12( xu−xl)=5−12( 1−(−2))=3(5−12)=1.8541

The two interior points are as follows:

First point is,

x1=xl+d=−2+1.8541=−0.1459

Second point is,

x2=xu−d=1−1.8541=−0.8541

Now, comparing the value of function at these interior points as shown below:

For x=−0.1459,

f(x1)=−x14−2x13−−8x12−5x1=−(−0.1459)4−2(−0.1459)3−8(−0.1459)2−5(−0.1459)=0.5650

For x=−0.8541,

f(x2)=−x24−2x23−8x22−5x2=−(−0.8541)4−2(−0.8541)3−8(−0.8541)2−5(−0.8541)=−0.8514

As f(x1)>f(x2),

Therefore, the maximum is in the interval defined by x2,x1 and xu

Where, x1 is the optimum.

The error at this point can be computed as follow:

εs=(1−R)| xu−xlxopt |⋅100, where R=5−12εs=(1−5−12)| 1−(−2)−0.1459 |⋅100εs=(1−5−12)(3−0.1459)⋅100εs=785.40%

Therefore, the domain of x to the left of x2 from −2 to x2 can be eliminated; because it does not contain the maximum value.

For this case, x2= −0.8541 becomes new xl for the next round.

Iteration 2: Here, xl=−0.8541 and xu=1. The first golden ratio is used to create two interior points as follow:

d=5−12( xu−xl)=5−12( 1−(−0.8541 ))=1.1459

The two new interior points are as follows:

First point is,

x1=xl+d=−0.8541 +1.1459=0.2918

Second point is,

x2=xu−d=1−1.1459=−0.1459

Now, comparing the value of function at these interior points as shown below:

For x=0.2918,

f(x1)=−x14−2x13−−8x12−5x1=−(0.2918)4−2(0.2918)3−8(0.2918)2−5(0.2918)=−2.197

For x=0.1459,

f(x2)=−x24−2x23−8x22−5x2=−(−0.1459)4−2(−0.1459)3−8(−0.1459)2−5(−0.1459)=0.5650

As f(x2)>f(x1)

For this case, x1=0.2918 become new xu, for the next round.

Now,

εs≃(1−R)|xu−xlxopt|⋅100,where R=5−12εs≃(1−5−12)|1−(−0.8541)−0.1459|⋅100εs≃(1−5−12)(1.8541−0.1459)⋅100εs≃485.40%

Iteration 3: Here, xl=−0.8541 and xu=0.2918. The first golden ratio is used to create two interior points as follow:

d=5−12( xu−xl)=5−12( 0.2918−(−0.8541))=0.7082

The two new interior points are as follows:

First interior point is,

x1=xl+d=−0.8541+0.7082=−0.1459

Second interior point is,

x2=xu−d=0.2918−0.7082=−0.4164

Now, comparing the value of function at these interior points as shown below:

For x=0.1459,

f(x1)=−x14−2x13−−8x12−5x1=−(−0.1459)4−2(−0.1459)3−8(−0.1459)2−5(−0.1459)=0.5650

For x=0.4164,

f(x2)=−x24−2x23−8x22−5x2=−(−0.4164)4−2(−0.4164)3−8(−0.4164)2−5(−0.4164)=0.8090

As f(x2)>f(x1)

Therefore, for this case, x1=−0.1459 becomes new xu for the next round.

Proceeding like this the iterations can be tabulated below as:

ixl−f(xl)x2f(x2)x1f(x1)xuf(xu)xoptεscomparison1−2−22−0.8541−0.8514−0.14590.56501−16−0.1459785.41f(x1)>f(x2)2−0.8541−0.8514−0.14590.56500.2918−2.1971−16−0.1459485.41f(x2)>f(x1)3−0.8541−0.8514−0.41640.8090−0.14590.56500.2918−2.197−0.4164105.11f(x2)>f(x1)4−0.8541−0.8514−0.58360.4750−0.41640.8090−0.14590.5650−0.416464.96f(x1)>f(x2)5−0.58360.4750−0.41640.8090−0.31310.8330−0.14590.5650−0.313153.40f(x1)>f(x2)6−0.41640.8090−0.31310.8330−0.24920.7760−0.14590.5650−0.313133f(x2)>f(x1)7−0.41640.8090−0.35250.8410−0.31310.8330−0.24920.7764−0.352518.11f(x2)>f(x1)8−0.41640.8090−0.37690.8351−0.35250.8410−0.31310.8330−0.352511.19f(x1)>f(x2)9−0.37690.8351−0.35250.8410−0.33750.8401−0.31310.8330−0.35256.92f(x2)>f(x1)10−0.37690.8351−0.36190.8393−0.35250.8410−0.33750.8401−0.35254.28f(x1)>f(x2)11−0.36190.8393−0.35250.8410−0.34680.8410−0.33750.8401−0.34682.69f(x1)>f(x2)12−0.35250.8410−0.34680.8410−0.34320.8412−0.33750.8401−0.34681.66f(x1)>f(x2)13−0.35250.8410−0.34900.8411−0.34680.8410−0.34320.8412−0.34681.03f(x1)>f(x2)14−0.34900.8411−0.34680.8410−0.34540.8411−0.3432−0.8412−0.34680.63f(x1)>f(x2)

Thus, the result converges to true value x=−0.3468.

(b)

Expert Solution

To determine

To calculate: The maximum of the function f(x)=−x4−2x3−8x2−5x with the help of parabolic interpolation. Take initial guesses as x0=−2,x1=−1,x2=1.

Answer to Problem 8P

Solution:

The maximum of the function f(x)=−x4−2x3−8x2−5x is, x=−0.3472 after 4 iterations.

Explanation of Solution

Given information:

The function f(x)=−x4−2x3−8x2−5x. Initial guesses x0=−2,x1=−1,x2=1 and the number of iterations is, 4.

Formula used:

Consider three points jointly bracket an optimum, thus a unique parabola through these three points can be determined. On differentiating and setting it equal to zero estimate of optimal can be computed.

Consider x0,x1 and x2 as initial guesses and x3 be the value of x that corresponds to the maximum value of quadratic which fit into guesses, then,

x3=f(x0)(x12−x22)+f(x1)(x22−x02)+f(x2)(x02−x12)2f(x0)(x1−x2)+2f(x1)(x2−x0)+2f(x2)(x0−x1) …… (1)

Calculation:

Consider function f(x), in standard form as:

f(x)=−x4−2x3−8x2−5x

With initial guesses x0=−2,x1=−1,x2=1.

Iteration 1: Function values at these three initial points is,

For x=−2,

f(x0)=f(−2)=−(−2)4−2(−2)3−8(−2)2−5(−2)=−22

For x=−1,

f(x1)=f(−1)=−(−1)4−2(−1)3−8(−1)2−5(−1)=−2

For x=1,

f(x2)=f(1)=−(1)4−2(1)3−8(1)2−5(1)=−16

Substitute the value of x0,x1,x2 and f(x0),f(x1),f(x2) in equation (1) to get value of x3,

x3=f(x0)(x12−x22)+f(x1)(x22−x02)+f(x2)(x02−x12)2f(x0)(x1−x2)+2f(x1)(x2−x0)+2f(x2)(x0−x1)=(−22)((−1)2−12)+(−2)(12−(−2)2)+(−16)((−2)2−(−1)2)2(−22)((−1)−1)+2(−2)(1−(−2))+2(−16)((−2)−(−1))=−0.3889

The value of function at x3 is,

f(x3)=f(−0.3889)=−(−0.3889)4−2(−0.3889)3−8(−0.3889)2−5(−0.3889)=0.8293

Therefore, f(x3)>f(x1) then the new value of x is to the right of intermediate point x1 and the lower guess x0 is discarded.

Iteration 2: Now the initial guesses are x0=−1,x1=−0.3889,x2=1

Function values at these three initial points are,

For x=−1,

f(x0)=f(−1)=−(−1)4−2(−1)3−8(−1)2−5(−1)=−2

For x=−0.3889,

f(x1)=f(−0.3889)=−(−0.3889)4−2(−0.3889)3−8(−0.3889)2−5(−0.3889)=0.8293

For x=1,

f(x2)=f(1)=−(1)4−2(1)3−8(1)2−5(1)=−16

Substitute the value of x0,x1,x2 and f(x0),f(x1),f(x2) in equation (1) to get value of x3,

x3=f(x0)(x12−x22)+f(x1)(x22−x02)+f(x2)(x02−x12)2f(x0)(x1−x2)+2f(x1)(x2−x0)+2f(x2)(x0−x1)=(−2)((−0.3889)2−12)+(0.8293)(12−(−1)2)+(−16)((−1)2−(−0.3889)2)2(−2)((−0.3889)−1)+2(0.8293)(1−(−1))+2(−16)((−1)−(−0.3889))=−0.4179

The value of the provided function at x3 is,

f(x3)=f(−0.4179)=−(−0.4179)4−2(−0.4179)3−8(−0.4179)2−5(−0.4179)=0.8077

Therefore, f(x1)>f(x3) then the new value of x is to the right of intermediate point x1 and the upper guess x2 is discarded.

Iteration 3: Now the initial guesses are x0=−1,x1=−0.4179,x2=−0.3889

Function values at these three initial points is,

For x=−1,

f(x0)=f(−1)=−(−1)4−2(−1)3−8(−1)2−5(−1)=−2

The function for x1

f(x1)=f(−0.4179)=−(−0.4179)4−2(−0.4179)3−8(−0.4179)2−5(−0.4179)=0.8077

And for x2,

f(x2)=f(−0.3889)=−(−0.3889)4−2(−0.3889)3−8(−0.3889)2−5(−0.3889)=0.8293

Substitute the value of x0,x1,x2 and f(x0),f(x1),f(x2) in equation (1) to get value of x3,

x3=f(x0)(x12−x22)+f(x1)(x22−x02)+f(x2)(x02−x12)2f(x0)(x1−x2)+2f(x1)(x2−x0)+2f(x2)(x0−x1)={(−2)((−0.4179)2−(−0.3889)2)+(0.8077)((−0.3889)2−(−1)2)+(0.8293)((−1)2−(−0.4179)2)}{2(−2)((−0.4179)−(−0.3889))+2(0.8077)((−0.3889)−(−1))+2(0.8293)((−1)−(−0.4179))}=−0.3476

And value of function at x3 is,

f(x3)=f(−0.3476)=−(−0.3476)4−2(−0.3476)3−8(−0.3476)2−5(−0.3476)=0.8408

Therefore, f(x3)>f(x1) then the new value of x is to the right of intermediate point x1 and the lower guess x0 is discarded.

Iteration 4: Now the initial guesses are x0=−0.4179,x1=−0.3889,x2=−0.3476

Function values at these three initial points is,

For x=−0.4179,

f(x0)=f(−0.4179)=−(−0.4179)4−2(−0.4179)3−8(−0.4179)2−5(−0.4179)=0.8077

For x=−0.3889,

f(x1)=f(−0.3889)=−(−0.3889)4−2(−0.3889)3−8(−0.3889)2−5(−0.3889)=0.8293

For x=−0.3476,

f(x2)=f(−0.3476)=−(−0.3476)4−2(−0.3476)3−8(−0.3476)2−5(−0.3476)=0.8408

Substitute the value of x0,x1,x2 and f(x0),f(x1),f(x2) in equation (1) to get value of x3,

x3=(f(x0)(x12−x22)+f(x1)(x22−x02)+f(x2)(x02−x12))(2f(x0)(x1−x2)+2f(x1)(x2−x0)+2f(x2)(x0−x1))=((0.8077)((−0.3889)2−(−0.3476)2)+(0.8293)((−0.3476)2−(−0.4179)2)+(0.8408)((−0.4179)2−(−0.3889)2))(2(0.8077)((−0.3889)−(−0.3476))+2(0.8293)((−0.3476)−(−0.4179))+2(0.8408)((−0.4179)−(−0.3889)))=−0.3472

And value of function at x3 is,

f(x3)=f(−0.3472)=−(−0.3472)4−2(−0.3472)3−8(−0.3472)2−5(−0.3472)=0.8408

Therefore, f(x1)>f(x3) then the new value of x is to the right of intermediate point x1 and the lower guess x0 is discarded.

And the process continues with a summary shown below in a table:

ix0f(x0)x1f(x1)x2f(x2)x3f(x3)εa(%)1−2−22−1−21−16−0.38890.82932−1−2−0.38890.82931−16−0.41790.80776.943−1−2−0.41790.8077−0.38890.8293−0.34760.840820.224−0.41790.8077−0.38890.8293−0.34760.8408−0.34720.84080.115

Thus, after four iterations result is converging to true value x=−0.3472.

(c)

Expert Solution

To determine

To calculate: The maximum of the function f(x)=−x4−2x3−8x2−5x with the help of Newton’s method. Take initial guess as x0=−1.

Answer to Problem 8P

Solution:

The maximum of the function f(x)=−x4−2x3−8x2−5x is, −0.3472.

Explanation of Solution

Given information:

The function f(x)=−x4−2x3−8x2−5x. Initial guess is x0=−1 and the acceptable error is, et=1%.

Formula used:

Newton Method is similar to Newton Raphson as it does not require initial guesses that bracket the optimum solution.

For any function f(x) optimal solution can be attained by performing iterations following the formula given below:

xi+1=xi−f′(xi)f″(xi) ...... (1)

Calculation:

Consider function f(x),

f(x)=−x4−2x3−8x2−5x

With initial guesses x0=−1,εt=1%.

First and second derivatives of function that is, f′(x) and f″(x), can be evaluated as:

f′(x)= −4x3−6x2−16x−5f″(x)=−12x2−12x−16

Iteration 1:Initially for x0=−1, values of first and second derivatives of function are as follows:

f′(x)= −4x3−6x2−16x−5f′(−1)=−4(−1)3−6(−1)2−16(−1)−5=9

For second derivative,

f″(x)=−12x2−12x−16f″(−1)=−12(−1)2−12(−1)−16=−16

Therefore, x1 is calculated using formula given as equation (2),

x1=x0−f′(x0)f″(x0)=−1−(9)(−16)=−0.4375

And

f(x1)=−x14−2x13−8x12−5x1=−(−0.4375)4−2(−0.4375)3−8(−0.4375)2−5(−0.4375)=0.7871

Iteration 2:Now for x1=−0.4375, values of first and second derivatives of function are as follows:

f′(x1)= −4x13−6x12−16x1−5f′(−0.4375)=−4(−0.4375)3−6(−0.4375)2−16(−0.4375)−5=1.1865

For second derivative,

f″(x)=−12x2−12x−16f″(−0.4375)=−12(−0.4375)2−12(−0.4375)−16=−13.045

Therefore, x2 is calculated using formula given as equation (2)

x2=x1−f′(x1)f″(x1)=−0.4375−(1.1865)(−13.045)=−0.3465

And

f(x2)=−x24−2x23−8x22−5x2=−(−0.3465)4−2(−0.3465)3−8(−0.3465)2−5(−0.3465)=0.8408

Iteration 3:Now for x2=−0.3465, values of first and second derivatives of function are as follows:

f′(x2)= −4x23−6x22−16x2−5f′(−0.3465)=−4(−0.3465)3−6(−0.3465)2−16(−0.3465)−5=−0.0099

For second derivative,

f″(x2)=−12x22−12x2−16f″(−0.3465)=−12(−0.3465)2−12(−0.3465)−16=−13.2827

Therefore, x3 is calculated using formula given as equation (2)

x3=x2−f′(x2)f″(x2)=−0.3465−(−0.0099)(−13.2827)=−0.3472

And

f(x3)=−x34−2x33−8x32−5x3=−(−0.3472)4−2(−0.3472)3−8(−0.3472)2−5(−0.3472)=0.8408

Iteration 4:Now for x3=−0.3472, values of first and second derivatives of function are as follows:

f′(x3)= −4x33−6x32−16x3−5f′(−0.3472)=−4(−0.3472)3−6(−0.3472)2−16(−0.3472)−5=−0.00067

For second derivative,

f″(x2)=−12x22−12x2−16f″(−0.3472)=−12(−0.3472)2−12(−0.3472)−16=−13.2801

Therefore, x4 is calculated using formula given as equation (2)

x4=x3−f′(x3)f″(x3)=−0.3472−(−0.00067)(−13.2801)=−0.347250

Maintaining the error percentage using equation (3) iterations can be summarized as shown in table below:

ixf(x)f′(x)f″(x)εa0−1−29−161−0.43750.78711.865−13.04616.162−0.34650.8408−0.0931−13.2850.113−0.34720.8408−0.0099−13.28000

Thus, within four iterations, the result converges to true value −0.3472.

Therefore, the maximum of the function f(x)=−x4−2x3−8x2−5x is −0.3472.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Students have asked these similar questions

1. Name the ongiewing) 2. Name five pairs of supple 3 27 and 19 form a angles 210 and 21 are complementary angies 4. m210=32 mal!= 5 mc11-72 m10= 6 m210-4x mc11=2x x= 7 m210=x m 11 =x+20; x= 12 and 213 are supplementary angles 8 ma 12 2y m13-3y-15 y= 9 m 12-y+10 m13-3y+ 10: y= 10. The measure of 212 is five times the measure of 13. Find the 213 and 214 are complementary angles, and 14 and 15 are supplementary angies 11 mc13 47 m/14- 12 m 14-78 m13- m215- m15 13 m15-135 m. 13- m.14

Complete solutions need handwriting. For all only sure experts solve it correct complete solutions

The graph below shows the U.S. federal expenses for 2012. A) estimate the fraction of the total expenses that were spent on Medicare. Write your answer as the closest fraction whose denominator is 100. B) estimate the fraction of the total expenses that were spent on Medicare and Medicaid. Write your answer as the closest fraction, whose denominator is 100.

Answer 2

Textbook Question

Chapter 13, Problem 8P

Employ the following methods to find the maximum of the function from Prob. 13.7:

(a) Golden-section search ( x l = − 2 , x u = 1 , ε s = 1 % ) .

(b) Parabolic interpolation ( x 0 = − 2 , x 1 = − 1 , x 2 = 1 , iterations = 4 ) . Select new points sequentially as in the secant method.

(c) Newton's method ( x 0 = − 1 , ε s = 1 % )

(a)

Expert Solution

To determine

To calculate: The maximum of the function f(x)=−x4−2x3−8x2−5x with the help of Golden-section search. Take initial guesses as xl=−2,xu=1, andan acceptable error as et=1%.

Answer to Problem 8P

Solution:

The maximum of the function f(x)=−x4−2x3−8x2−5x is, x=−0.3468 after 14 iterations.

Explanation of Solution

Given information:

The function f(x)=−x4−2x3−8x2−5x. Initial guesses xl=−2,xu=1 and value of acceptable error is et=1%.

Formula used:

The golden-search method with two initial guesses, xl and xu that encloses one local extreme of function f(x). Two interior points x1 and x2 with the golden ratio d is given by,

d=5−12( xu−xl)x1=xl+dx2=xu−d

Evaluated the function at the above two interior points. Two results can occur,

If f(x1)>f(x2), then x2 becomes the new xl for the next iteration.

If f(x2)>f(x1), then x1 becomes the new xu for the next iteration.

Calculation:

Consider function f(x), in standard form as

f(x)=−x4−2x3−8x2−5x

With xl=−2 , xu=1 and εs=1%

Iteration 1: First golden ratio is used to create two interior points as,

d=5−12( xu−xl)=5−12( 1−(−2))=3(5−12)=1.8541

The two interior points are as follows:

First point is,

x1=xl+d=−2+1.8541=−0.1459

Second point is,

x2=xu−d=1−1.8541=−0.8541

Now, comparing the value of function at these interior points as shown below:

For x=−0.1459,

f(x1)=−x14−2x13−−8x12−5x1=−(−0.1459)4−2(−0.1459)3−8(−0.1459)2−5(−0.1459)=0.5650

For x=−0.8541,

f(x2)=−x24−2x23−8x22−5x2=−(−0.8541)4−2(−0.8541)3−8(−0.8541)2−5(−0.8541)=−0.8514

As f(x1)>f(x2),

Therefore, the maximum is in the interval defined by x2,x1 and xu

Where, x1 is the optimum.

The error at this point can be computed as follow:

εs=(1−R)| xu−xlxopt |⋅100, where R=5−12εs=(1−5−12)| 1−(−2)−0.1459 |⋅100εs=(1−5−12)(3−0.1459)⋅100εs=785.40%

Therefore, the domain of x to the left of x2 from −2 to x2 can be eliminated; because it does not contain the maximum value.

For this case, x2= −0.8541 becomes new xl for the next round.

Iteration 2: Here, xl=−0.8541 and xu=1. The first golden ratio is used to create two interior points as follow:

d=5−12( xu−xl)=5−12( 1−(−0.8541 ))=1.1459

The two new interior points are as follows:

First point is,

x1=xl+d=−0.8541 +1.1459=0.2918

Second point is,

x2=xu−d=1−1.1459=−0.1459

Now, comparing the value of function at these interior points as shown below:

For x=0.2918,

f(x1)=−x14−2x13−−8x12−5x1=−(0.2918)4−2(0.2918)3−8(0.2918)2−5(0.2918)=−2.197

For x=0.1459,

f(x2)=−x24−2x23−8x22−5x2=−(−0.1459)4−2(−0.1459)3−8(−0.1459)2−5(−0.1459)=0.5650

As f(x2)>f(x1)

For this case, x1=0.2918 become new xu, for the next round.

Now,

εs≃(1−R)|xu−xlxopt|⋅100,where R=5−12εs≃(1−5−12)|1−(−0.8541)−0.1459|⋅100εs≃(1−5−12)(1.8541−0.1459)⋅100εs≃485.40%

Iteration 3: Here, xl=−0.8541 and xu=0.2918. The first golden ratio is used to create two interior points as follow:

d=5−12( xu−xl)=5−12( 0.2918−(−0.8541))=0.7082

The two new interior points are as follows:

First interior point is,

x1=xl+d=−0.8541+0.7082=−0.1459

Second interior point is,

x2=xu−d=0.2918−0.7082=−0.4164

Now, comparing the value of function at these interior points as shown below:

For x=0.1459,

f(x1)=−x14−2x13−−8x12−5x1=−(−0.1459)4−2(−0.1459)3−8(−0.1459)2−5(−0.1459)=0.5650

For x=0.4164,

f(x2)=−x24−2x23−8x22−5x2=−(−0.4164)4−2(−0.4164)3−8(−0.4164)2−5(−0.4164)=0.8090

As f(x2)>f(x1)

Therefore, for this case, x1=−0.1459 becomes new xu for the next round.

Proceeding like this the iterations can be tabulated below as:

ixl−f(xl)x2f(x2)x1f(x1)xuf(xu)xoptεscomparison1−2−22−0.8541−0.8514−0.14590.56501−16−0.1459785.41f(x1)>f(x2)2−0.8541−0.8514−0.14590.56500.2918−2.1971−16−0.1459485.41f(x2)>f(x1)3−0.8541−0.8514−0.41640.8090−0.14590.56500.2918−2.197−0.4164105.11f(x2)>f(x1)4−0.8541−0.8514−0.58360.4750−0.41640.8090−0.14590.5650−0.416464.96f(x1)>f(x2)5−0.58360.4750−0.41640.8090−0.31310.8330−0.14590.5650−0.313153.40f(x1)>f(x2)6−0.41640.8090−0.31310.8330−0.24920.7760−0.14590.5650−0.313133f(x2)>f(x1)7−0.41640.8090−0.35250.8410−0.31310.8330−0.24920.7764−0.352518.11f(x2)>f(x1)8−0.41640.8090−0.37690.8351−0.35250.8410−0.31310.8330−0.352511.19f(x1)>f(x2)9−0.37690.8351−0.35250.8410−0.33750.8401−0.31310.8330−0.35256.92f(x2)>f(x1)10−0.37690.8351−0.36190.8393−0.35250.8410−0.33750.8401−0.35254.28f(x1)>f(x2)11−0.36190.8393−0.35250.8410−0.34680.8410−0.33750.8401−0.34682.69f(x1)>f(x2)12−0.35250.8410−0.34680.8410−0.34320.8412−0.33750.8401−0.34681.66f(x1)>f(x2)13−0.35250.8410−0.34900.8411−0.34680.8410−0.34320.8412−0.34681.03f(x1)>f(x2)14−0.34900.8411−0.34680.8410−0.34540.8411−0.3432−0.8412−0.34680.63f(x1)>f(x2)

Thus, the result converges to true value x=−0.3468.

(b)

Expert Solution

To determine

To calculate: The maximum of the function f(x)=−x4−2x3−8x2−5x with the help of parabolic interpolation. Take initial guesses as x0=−2,x1=−1,x2=1.

Answer to Problem 8P

Solution:

The maximum of the function f(x)=−x4−2x3−8x2−5x is, x=−0.3472 after 4 iterations.

Explanation of Solution

Given information:

The function f(x)=−x4−2x3−8x2−5x. Initial guesses x0=−2,x1=−1,x2=1 and the number of iterations is, 4.

Formula used:

Consider three points jointly bracket an optimum, thus a unique parabola through these three points can be determined. On differentiating and setting it equal to zero estimate of optimal can be computed.

Consider x0,x1 and x2 as initial guesses and x3 be the value of x that corresponds to the maximum value of quadratic which fit into guesses, then,

x3=f(x0)(x12−x22)+f(x1)(x22−x02)+f(x2)(x02−x12)2f(x0)(x1−x2)+2f(x1)(x2−x0)+2f(x2)(x0−x1) …… (1)

Calculation:

Consider function f(x), in standard form as:

f(x)=−x4−2x3−8x2−5x

With initial guesses x0=−2,x1=−1,x2=1.

Iteration 1: Function values at these three initial points is,

For x=−2,

f(x0)=f(−2)=−(−2)4−2(−2)3−8(−2)2−5(−2)=−22

For x=−1,

f(x1)=f(−1)=−(−1)4−2(−1)3−8(−1)2−5(−1)=−2

For x=1,

f(x2)=f(1)=−(1)4−2(1)3−8(1)2−5(1)=−16

Substitute the value of x0,x1,x2 and f(x0),f(x1),f(x2) in equation (1) to get value of x3,

x3=f(x0)(x12−x22)+f(x1)(x22−x02)+f(x2)(x02−x12)2f(x0)(x1−x2)+2f(x1)(x2−x0)+2f(x2)(x0−x1)=(−22)((−1)2−12)+(−2)(12−(−2)2)+(−16)((−2)2−(−1)2)2(−22)((−1)−1)+2(−2)(1−(−2))+2(−16)((−2)−(−1))=−0.3889

The value of function at x3 is,

f(x3)=f(−0.3889)=−(−0.3889)4−2(−0.3889)3−8(−0.3889)2−5(−0.3889)=0.8293

Therefore, f(x3)>f(x1) then the new value of x is to the right of intermediate point x1 and the lower guess x0 is discarded.

Iteration 2: Now the initial guesses are x0=−1,x1=−0.3889,x2=1

Function values at these three initial points are,

For x=−1,

f(x0)=f(−1)=−(−1)4−2(−1)3−8(−1)2−5(−1)=−2

For x=−0.3889,

f(x1)=f(−0.3889)=−(−0.3889)4−2(−0.3889)3−8(−0.3889)2−5(−0.3889)=0.8293

For x=1,

f(x2)=f(1)=−(1)4−2(1)3−8(1)2−5(1)=−16

Substitute the value of x0,x1,x2 and f(x0),f(x1),f(x2) in equation (1) to get value of x3,

x3=f(x0)(x12−x22)+f(x1)(x22−x02)+f(x2)(x02−x12)2f(x0)(x1−x2)+2f(x1)(x2−x0)+2f(x2)(x0−x1)=(−2)((−0.3889)2−12)+(0.8293)(12−(−1)2)+(−16)((−1)2−(−0.3889)2)2(−2)((−0.3889)−1)+2(0.8293)(1−(−1))+2(−16)((−1)−(−0.3889))=−0.4179

The value of the provided function at x3 is,

f(x3)=f(−0.4179)=−(−0.4179)4−2(−0.4179)3−8(−0.4179)2−5(−0.4179)=0.8077

Therefore, f(x1)>f(x3) then the new value of x is to the right of intermediate point x1 and the upper guess x2 is discarded.

Iteration 3: Now the initial guesses are x0=−1,x1=−0.4179,x2=−0.3889

Function values at these three initial points is,

For x=−1,

f(x0)=f(−1)=−(−1)4−2(−1)3−8(−1)2−5(−1)=−2

The function for x1

f(x1)=f(−0.4179)=−(−0.4179)4−2(−0.4179)3−8(−0.4179)2−5(−0.4179)=0.8077

And for x2,

f(x2)=f(−0.3889)=−(−0.3889)4−2(−0.3889)3−8(−0.3889)2−5(−0.3889)=0.8293

Substitute the value of x0,x1,x2 and f(x0),f(x1),f(x2) in equation (1) to get value of x3,

x3=f(x0)(x12−x22)+f(x1)(x22−x02)+f(x2)(x02−x12)2f(x0)(x1−x2)+2f(x1)(x2−x0)+2f(x2)(x0−x1)={(−2)((−0.4179)2−(−0.3889)2)+(0.8077)((−0.3889)2−(−1)2)+(0.8293)((−1)2−(−0.4179)2)}{2(−2)((−0.4179)−(−0.3889))+2(0.8077)((−0.3889)−(−1))+2(0.8293)((−1)−(−0.4179))}=−0.3476

And value of function at x3 is,

f(x3)=f(−0.3476)=−(−0.3476)4−2(−0.3476)3−8(−0.3476)2−5(−0.3476)=0.8408

Therefore, f(x3)>f(x1) then the new value of x is to the right of intermediate point x1 and the lower guess x0 is discarded.

Iteration 4: Now the initial guesses are x0=−0.4179,x1=−0.3889,x2=−0.3476

Function values at these three initial points is,

For x=−0.4179,

f(x0)=f(−0.4179)=−(−0.4179)4−2(−0.4179)3−8(−0.4179)2−5(−0.4179)=0.8077

For x=−0.3889,

f(x1)=f(−0.3889)=−(−0.3889)4−2(−0.3889)3−8(−0.3889)2−5(−0.3889)=0.8293

For x=−0.3476,

f(x2)=f(−0.3476)=−(−0.3476)4−2(−0.3476)3−8(−0.3476)2−5(−0.3476)=0.8408

Substitute the value of x0,x1,x2 and f(x0),f(x1),f(x2) in equation (1) to get value of x3,

x3=(f(x0)(x12−x22)+f(x1)(x22−x02)+f(x2)(x02−x12))(2f(x0)(x1−x2)+2f(x1)(x2−x0)+2f(x2)(x0−x1))=((0.8077)((−0.3889)2−(−0.3476)2)+(0.8293)((−0.3476)2−(−0.4179)2)+(0.8408)((−0.4179)2−(−0.3889)2))(2(0.8077)((−0.3889)−(−0.3476))+2(0.8293)((−0.3476)−(−0.4179))+2(0.8408)((−0.4179)−(−0.3889)))=−0.3472

And value of function at x3 is,

f(x3)=f(−0.3472)=−(−0.3472)4−2(−0.3472)3−8(−0.3472)2−5(−0.3472)=0.8408

Therefore, f(x1)>f(x3) then the new value of x is to the right of intermediate point x1 and the lower guess x0 is discarded.

And the process continues with a summary shown below in a table:

ix0f(x0)x1f(x1)x2f(x2)x3f(x3)εa(%)1−2−22−1−21−16−0.38890.82932−1−2−0.38890.82931−16−0.41790.80776.943−1−2−0.41790.8077−0.38890.8293−0.34760.840820.224−0.41790.8077−0.38890.8293−0.34760.8408−0.34720.84080.115

Thus, after four iterations result is converging to true value x=−0.3472.

(c)

Expert Solution

To determine

To calculate: The maximum of the function f(x)=−x4−2x3−8x2−5x with the help of Newton’s method. Take initial guess as x0=−1.

Answer to Problem 8P

Solution:

The maximum of the function f(x)=−x4−2x3−8x2−5x is, −0.3472.

Explanation of Solution

Given information:

The function f(x)=−x4−2x3−8x2−5x. Initial guess is x0=−1 and the acceptable error is, et=1%.

Formula used:

Newton Method is similar to Newton Raphson as it does not require initial guesses that bracket the optimum solution.

For any function f(x) optimal solution can be attained by performing iterations following the formula given below:

xi+1=xi−f′(xi)f″(xi) ...... (1)

Calculation:

Consider function f(x),

f(x)=−x4−2x3−8x2−5x

With initial guesses x0=−1,εt=1%.

First and second derivatives of function that is, f′(x) and f″(x), can be evaluated as:

f′(x)= −4x3−6x2−16x−5f″(x)=−12x2−12x−16

Iteration 1:Initially for x0=−1, values of first and second derivatives of function are as follows:

f′(x)= −4x3−6x2−16x−5f′(−1)=−4(−1)3−6(−1)2−16(−1)−5=9

For second derivative,

f″(x)=−12x2−12x−16f″(−1)=−12(−1)2−12(−1)−16=−16

Therefore, x1 is calculated using formula given as equation (2),

x1=x0−f′(x0)f″(x0)=−1−(9)(−16)=−0.4375

And

f(x1)=−x14−2x13−8x12−5x1=−(−0.4375)4−2(−0.4375)3−8(−0.4375)2−5(−0.4375)=0.7871

Iteration 2:Now for x1=−0.4375, values of first and second derivatives of function are as follows:

f′(x1)= −4x13−6x12−16x1−5f′(−0.4375)=−4(−0.4375)3−6(−0.4375)2−16(−0.4375)−5=1.1865

For second derivative,

f″(x)=−12x2−12x−16f″(−0.4375)=−12(−0.4375)2−12(−0.4375)−16=−13.045

Therefore, x2 is calculated using formula given as equation (2)

x2=x1−f′(x1)f″(x1)=−0.4375−(1.1865)(−13.045)=−0.3465

And

f(x2)=−x24−2x23−8x22−5x2=−(−0.3465)4−2(−0.3465)3−8(−0.3465)2−5(−0.3465)=0.8408

Iteration 3:Now for x2=−0.3465, values of first and second derivatives of function are as follows:

f′(x2)= −4x23−6x22−16x2−5f′(−0.3465)=−4(−0.3465)3−6(−0.3465)2−16(−0.3465)−5=−0.0099

For second derivative,

f″(x2)=−12x22−12x2−16f″(−0.3465)=−12(−0.3465)2−12(−0.3465)−16=−13.2827

Therefore, x3 is calculated using formula given as equation (2)

x3=x2−f′(x2)f″(x2)=−0.3465−(−0.0099)(−13.2827)=−0.3472

And

f(x3)=−x34−2x33−8x32−5x3=−(−0.3472)4−2(−0.3472)3−8(−0.3472)2−5(−0.3472)=0.8408

Iteration 4:Now for x3=−0.3472, values of first and second derivatives of function are as follows:

f′(x3)= −4x33−6x32−16x3−5f′(−0.3472)=−4(−0.3472)3−6(−0.3472)2−16(−0.3472)−5=−0.00067

For second derivative,

f″(x2)=−12x22−12x2−16f″(−0.3472)=−12(−0.3472)2−12(−0.3472)−16=−13.2801

Therefore, x4 is calculated using formula given as equation (2)

x4=x3−f′(x3)f″(x3)=−0.3472−(−0.00067)(−13.2801)=−0.347250

Maintaining the error percentage using equation (3) iterations can be summarized as shown in table below:

ixf(x)f′(x)f″(x)εa0−1−29−161−0.43750.78711.865−13.04616.162−0.34650.8408−0.0931−13.2850.113−0.34720.8408−0.0099−13.28000

Thus, within four iterations, the result converges to true value −0.3472.

Therefore, the maximum of the function f(x)=−x4−2x3−8x2−5x is −0.3472.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 3

Textbook Question

Chapter 13, Problem 8P

Employ the following methods to find the maximum of the function from Prob. 13.7:

(a) Golden-section search ( x l = − 2 , x u = 1 , ε s = 1 % ) .

(b) Parabolic interpolation ( x 0 = − 2 , x 1 = − 1 , x 2 = 1 , iterations = 4 ) . Select new points sequentially as in the secant method.

(c) Newton's method ( x 0 = − 1 , ε s = 1 % )

(a)

Expert Solution

To determine

To calculate: The maximum of the function f(x)=−x4−2x3−8x2−5x with the help of Golden-section search. Take initial guesses as xl=−2,xu=1, andan acceptable error as et=1%.

Answer to Problem 8P

Solution:

The maximum of the function f(x)=−x4−2x3−8x2−5x is, x=−0.3468 after 14 iterations.

Explanation of Solution

Given information:

The function f(x)=−x4−2x3−8x2−5x. Initial guesses xl=−2,xu=1 and value of acceptable error is et=1%.

Formula used:

The golden-search method with two initial guesses, xl and xu that encloses one local extreme of function f(x). Two interior points x1 and x2 with the golden ratio d is given by,

d=5−12( xu−xl)x1=xl+dx2=xu−d

Evaluated the function at the above two interior points. Two results can occur,

If f(x1)>f(x2), then x2 becomes the new xl for the next iteration.

If f(x2)>f(x1), then x1 becomes the new xu for the next iteration.

Calculation:

Consider function f(x), in standard form as

f(x)=−x4−2x3−8x2−5x

With xl=−2 , xu=1 and εs=1%

Iteration 1: First golden ratio is used to create two interior points as,

d=5−12( xu−xl)=5−12( 1−(−2))=3(5−12)=1.8541

The two interior points are as follows:

First point is,

x1=xl+d=−2+1.8541=−0.1459

Second point is,

x2=xu−d=1−1.8541=−0.8541

Now, comparing the value of function at these interior points as shown below:

For x=−0.1459,

f(x1)=−x14−2x13−−8x12−5x1=−(−0.1459)4−2(−0.1459)3−8(−0.1459)2−5(−0.1459)=0.5650

For x=−0.8541,

f(x2)=−x24−2x23−8x22−5x2=−(−0.8541)4−2(−0.8541)3−8(−0.8541)2−5(−0.8541)=−0.8514

As f(x1)>f(x2),

Therefore, the maximum is in the interval defined by x2,x1 and xu

Where, x1 is the optimum.

The error at this point can be computed as follow:

εs=(1−R)| xu−xlxopt |⋅100, where R=5−12εs=(1−5−12)| 1−(−2)−0.1459 |⋅100εs=(1−5−12)(3−0.1459)⋅100εs=785.40%

Therefore, the domain of x to the left of x2 from −2 to x2 can be eliminated; because it does not contain the maximum value.

For this case, x2= −0.8541 becomes new xl for the next round.

Iteration 2: Here, xl=−0.8541 and xu=1. The first golden ratio is used to create two interior points as follow:

d=5−12( xu−xl)=5−12( 1−(−0.8541 ))=1.1459

The two new interior points are as follows:

First point is,

x1=xl+d=−0.8541 +1.1459=0.2918

Second point is,

x2=xu−d=1−1.1459=−0.1459

Now, comparing the value of function at these interior points as shown below:

For x=0.2918,

f(x1)=−x14−2x13−−8x12−5x1=−(0.2918)4−2(0.2918)3−8(0.2918)2−5(0.2918)=−2.197

For x=0.1459,

f(x2)=−x24−2x23−8x22−5x2=−(−0.1459)4−2(−0.1459)3−8(−0.1459)2−5(−0.1459)=0.5650

As f(x2)>f(x1)

For this case, x1=0.2918 become new xu, for the next round.

Now,

εs≃(1−R)|xu−xlxopt|⋅100,where R=5−12εs≃(1−5−12)|1−(−0.8541)−0.1459|⋅100εs≃(1−5−12)(1.8541−0.1459)⋅100εs≃485.40%

Iteration 3: Here, xl=−0.8541 and xu=0.2918. The first golden ratio is used to create two interior points as follow:

d=5−12( xu−xl)=5−12( 0.2918−(−0.8541))=0.7082

The two new interior points are as follows:

First interior point is,

x1=xl+d=−0.8541+0.7082=−0.1459

Second interior point is,

x2=xu−d=0.2918−0.7082=−0.4164

Now, comparing the value of function at these interior points as shown below:

For x=0.1459,

f(x1)=−x14−2x13−−8x12−5x1=−(−0.1459)4−2(−0.1459)3−8(−0.1459)2−5(−0.1459)=0.5650

For x=0.4164,

f(x2)=−x24−2x23−8x22−5x2=−(−0.4164)4−2(−0.4164)3−8(−0.4164)2−5(−0.4164)=0.8090

As f(x2)>f(x1)

Therefore, for this case, x1=−0.1459 becomes new xu for the next round.

Proceeding like this the iterations can be tabulated below as:

ixl−f(xl)x2f(x2)x1f(x1)xuf(xu)xoptεscomparison1−2−22−0.8541−0.8514−0.14590.56501−16−0.1459785.41f(x1)>f(x2)2−0.8541−0.8514−0.14590.56500.2918−2.1971−16−0.1459485.41f(x2)>f(x1)3−0.8541−0.8514−0.41640.8090−0.14590.56500.2918−2.197−0.4164105.11f(x2)>f(x1)4−0.8541−0.8514−0.58360.4750−0.41640.8090−0.14590.5650−0.416464.96f(x1)>f(x2)5−0.58360.4750−0.41640.8090−0.31310.8330−0.14590.5650−0.313153.40f(x1)>f(x2)6−0.41640.8090−0.31310.8330−0.24920.7760−0.14590.5650−0.313133f(x2)>f(x1)7−0.41640.8090−0.35250.8410−0.31310.8330−0.24920.7764−0.352518.11f(x2)>f(x1)8−0.41640.8090−0.37690.8351−0.35250.8410−0.31310.8330−0.352511.19f(x1)>f(x2)9−0.37690.8351−0.35250.8410−0.33750.8401−0.31310.8330−0.35256.92f(x2)>f(x1)10−0.37690.8351−0.36190.8393−0.35250.8410−0.33750.8401−0.35254.28f(x1)>f(x2)11−0.36190.8393−0.35250.8410−0.34680.8410−0.33750.8401−0.34682.69f(x1)>f(x2)12−0.35250.8410−0.34680.8410−0.34320.8412−0.33750.8401−0.34681.66f(x1)>f(x2)13−0.35250.8410−0.34900.8411−0.34680.8410−0.34320.8412−0.34681.03f(x1)>f(x2)14−0.34900.8411−0.34680.8410−0.34540.8411−0.3432−0.8412−0.34680.63f(x1)>f(x2)

Thus, the result converges to true value x=−0.3468.

(b)

Expert Solution

To determine

To calculate: The maximum of the function f(x)=−x4−2x3−8x2−5x with the help of parabolic interpolation. Take initial guesses as x0=−2,x1=−1,x2=1.

Answer to Problem 8P

Solution:

The maximum of the function f(x)=−x4−2x3−8x2−5x is, x=−0.3472 after 4 iterations.

Explanation of Solution

Given information:

The function f(x)=−x4−2x3−8x2−5x. Initial guesses x0=−2,x1=−1,x2=1 and the number of iterations is, 4.

Formula used:

Consider three points jointly bracket an optimum, thus a unique parabola through these three points can be determined. On differentiating and setting it equal to zero estimate of optimal can be computed.

Consider x0,x1 and x2 as initial guesses and x3 be the value of x that corresponds to the maximum value of quadratic which fit into guesses, then,

x3=f(x0)(x12−x22)+f(x1)(x22−x02)+f(x2)(x02−x12)2f(x0)(x1−x2)+2f(x1)(x2−x0)+2f(x2)(x0−x1) …… (1)

Calculation:

Consider function f(x), in standard form as:

f(x)=−x4−2x3−8x2−5x

With initial guesses x0=−2,x1=−1,x2=1.

Iteration 1: Function values at these three initial points is,

For x=−2,

f(x0)=f(−2)=−(−2)4−2(−2)3−8(−2)2−5(−2)=−22

For x=−1,

f(x1)=f(−1)=−(−1)4−2(−1)3−8(−1)2−5(−1)=−2

For x=1,

f(x2)=f(1)=−(1)4−2(1)3−8(1)2−5(1)=−16

Substitute the value of x0,x1,x2 and f(x0),f(x1),f(x2) in equation (1) to get value of x3,

x3=f(x0)(x12−x22)+f(x1)(x22−x02)+f(x2)(x02−x12)2f(x0)(x1−x2)+2f(x1)(x2−x0)+2f(x2)(x0−x1)=(−22)((−1)2−12)+(−2)(12−(−2)2)+(−16)((−2)2−(−1)2)2(−22)((−1)−1)+2(−2)(1−(−2))+2(−16)((−2)−(−1))=−0.3889

The value of function at x3 is,

f(x3)=f(−0.3889)=−(−0.3889)4−2(−0.3889)3−8(−0.3889)2−5(−0.3889)=0.8293

Therefore, f(x3)>f(x1) then the new value of x is to the right of intermediate point x1 and the lower guess x0 is discarded.

Iteration 2: Now the initial guesses are x0=−1,x1=−0.3889,x2=1

Function values at these three initial points are,

For x=−1,

f(x0)=f(−1)=−(−1)4−2(−1)3−8(−1)2−5(−1)=−2

For x=−0.3889,

f(x1)=f(−0.3889)=−(−0.3889)4−2(−0.3889)3−8(−0.3889)2−5(−0.3889)=0.8293

For x=1,

f(x2)=f(1)=−(1)4−2(1)3−8(1)2−5(1)=−16

Substitute the value of x0,x1,x2 and f(x0),f(x1),f(x2) in equation (1) to get value of x3,

x3=f(x0)(x12−x22)+f(x1)(x22−x02)+f(x2)(x02−x12)2f(x0)(x1−x2)+2f(x1)(x2−x0)+2f(x2)(x0−x1)=(−2)((−0.3889)2−12)+(0.8293)(12−(−1)2)+(−16)((−1)2−(−0.3889)2)2(−2)((−0.3889)−1)+2(0.8293)(1−(−1))+2(−16)((−1)−(−0.3889))=−0.4179

The value of the provided function at x3 is,

f(x3)=f(−0.4179)=−(−0.4179)4−2(−0.4179)3−8(−0.4179)2−5(−0.4179)=0.8077

Therefore, f(x1)>f(x3) then the new value of x is to the right of intermediate point x1 and the upper guess x2 is discarded.

Iteration 3: Now the initial guesses are x0=−1,x1=−0.4179,x2=−0.3889

Function values at these three initial points is,

For x=−1,

f(x0)=f(−1)=−(−1)4−2(−1)3−8(−1)2−5(−1)=−2

The function for x1

f(x1)=f(−0.4179)=−(−0.4179)4−2(−0.4179)3−8(−0.4179)2−5(−0.4179)=0.8077

And for x2,

f(x2)=f(−0.3889)=−(−0.3889)4−2(−0.3889)3−8(−0.3889)2−5(−0.3889)=0.8293

Substitute the value of x0,x1,x2 and f(x0),f(x1),f(x2) in equation (1) to get value of x3,

x3=f(x0)(x12−x22)+f(x1)(x22−x02)+f(x2)(x02−x12)2f(x0)(x1−x2)+2f(x1)(x2−x0)+2f(x2)(x0−x1)={(−2)((−0.4179)2−(−0.3889)2)+(0.8077)((−0.3889)2−(−1)2)+(0.8293)((−1)2−(−0.4179)2)}{2(−2)((−0.4179)−(−0.3889))+2(0.8077)((−0.3889)−(−1))+2(0.8293)((−1)−(−0.4179))}=−0.3476

And value of function at x3 is,

f(x3)=f(−0.3476)=−(−0.3476)4−2(−0.3476)3−8(−0.3476)2−5(−0.3476)=0.8408

Therefore, f(x3)>f(x1) then the new value of x is to the right of intermediate point x1 and the lower guess x0 is discarded.

Iteration 4: Now the initial guesses are x0=−0.4179,x1=−0.3889,x2=−0.3476

Function values at these three initial points is,

For x=−0.4179,

f(x0)=f(−0.4179)=−(−0.4179)4−2(−0.4179)3−8(−0.4179)2−5(−0.4179)=0.8077

For x=−0.3889,

f(x1)=f(−0.3889)=−(−0.3889)4−2(−0.3889)3−8(−0.3889)2−5(−0.3889)=0.8293

For x=−0.3476,

f(x2)=f(−0.3476)=−(−0.3476)4−2(−0.3476)3−8(−0.3476)2−5(−0.3476)=0.8408

Substitute the value of x0,x1,x2 and f(x0),f(x1),f(x2) in equation (1) to get value of x3,

x3=(f(x0)(x12−x22)+f(x1)(x22−x02)+f(x2)(x02−x12))(2f(x0)(x1−x2)+2f(x1)(x2−x0)+2f(x2)(x0−x1))=((0.8077)((−0.3889)2−(−0.3476)2)+(0.8293)((−0.3476)2−(−0.4179)2)+(0.8408)((−0.4179)2−(−0.3889)2))(2(0.8077)((−0.3889)−(−0.3476))+2(0.8293)((−0.3476)−(−0.4179))+2(0.8408)((−0.4179)−(−0.3889)))=−0.3472

And value of function at x3 is,

f(x3)=f(−0.3472)=−(−0.3472)4−2(−0.3472)3−8(−0.3472)2−5(−0.3472)=0.8408

Therefore, f(x1)>f(x3) then the new value of x is to the right of intermediate point x1 and the lower guess x0 is discarded.

And the process continues with a summary shown below in a table:

ix0f(x0)x1f(x1)x2f(x2)x3f(x3)εa(%)1−2−22−1−21−16−0.38890.82932−1−2−0.38890.82931−16−0.41790.80776.943−1−2−0.41790.8077−0.38890.8293−0.34760.840820.224−0.41790.8077−0.38890.8293−0.34760.8408−0.34720.84080.115

Thus, after four iterations result is converging to true value x=−0.3472.

(c)

Expert Solution

To determine

To calculate: The maximum of the function f(x)=−x4−2x3−8x2−5x with the help of Newton’s method. Take initial guess as x0=−1.

Answer to Problem 8P

Solution:

The maximum of the function f(x)=−x4−2x3−8x2−5x is, −0.3472.

Explanation of Solution

Given information:

The function f(x)=−x4−2x3−8x2−5x. Initial guess is x0=−1 and the acceptable error is, et=1%.

Formula used:

Newton Method is similar to Newton Raphson as it does not require initial guesses that bracket the optimum solution.

For any function f(x) optimal solution can be attained by performing iterations following the formula given below:

xi+1=xi−f′(xi)f″(xi) ...... (1)

Calculation:

Consider function f(x),

f(x)=−x4−2x3−8x2−5x

With initial guesses x0=−1,εt=1%.

First and second derivatives of function that is, f′(x) and f″(x), can be evaluated as:

f′(x)= −4x3−6x2−16x−5f″(x)=−12x2−12x−16

Iteration 1:Initially for x0=−1, values of first and second derivatives of function are as follows:

f′(x)= −4x3−6x2−16x−5f′(−1)=−4(−1)3−6(−1)2−16(−1)−5=9

For second derivative,

f″(x)=−12x2−12x−16f″(−1)=−12(−1)2−12(−1)−16=−16

Therefore, x1 is calculated using formula given as equation (2),

x1=x0−f′(x0)f″(x0)=−1−(9)(−16)=−0.4375

And

f(x1)=−x14−2x13−8x12−5x1=−(−0.4375)4−2(−0.4375)3−8(−0.4375)2−5(−0.4375)=0.7871

Iteration 2:Now for x1=−0.4375, values of first and second derivatives of function are as follows:

f′(x1)= −4x13−6x12−16x1−5f′(−0.4375)=−4(−0.4375)3−6(−0.4375)2−16(−0.4375)−5=1.1865

For second derivative,

f″(x)=−12x2−12x−16f″(−0.4375)=−12(−0.4375)2−12(−0.4375)−16=−13.045

Therefore, x2 is calculated using formula given as equation (2)

x2=x1−f′(x1)f″(x1)=−0.4375−(1.1865)(−13.045)=−0.3465

And

f(x2)=−x24−2x23−8x22−5x2=−(−0.3465)4−2(−0.3465)3−8(−0.3465)2−5(−0.3465)=0.8408

Iteration 3:Now for x2=−0.3465, values of first and second derivatives of function are as follows:

f′(x2)= −4x23−6x22−16x2−5f′(−0.3465)=−4(−0.3465)3−6(−0.3465)2−16(−0.3465)−5=−0.0099

For second derivative,

f″(x2)=−12x22−12x2−16f″(−0.3465)=−12(−0.3465)2−12(−0.3465)−16=−13.2827

Therefore, x3 is calculated using formula given as equation (2)

x3=x2−f′(x2)f″(x2)=−0.3465−(−0.0099)(−13.2827)=−0.3472

And

f(x3)=−x34−2x33−8x32−5x3=−(−0.3472)4−2(−0.3472)3−8(−0.3472)2−5(−0.3472)=0.8408

Iteration 4:Now for x3=−0.3472, values of first and second derivatives of function are as follows:

f′(x3)= −4x33−6x32−16x3−5f′(−0.3472)=−4(−0.3472)3−6(−0.3472)2−16(−0.3472)−5=−0.00067

For second derivative,

f″(x2)=−12x22−12x2−16f″(−0.3472)=−12(−0.3472)2−12(−0.3472)−16=−13.2801

Therefore, x4 is calculated using formula given as equation (2)

x4=x3−f′(x3)f″(x3)=−0.3472−(−0.00067)(−13.2801)=−0.347250

Maintaining the error percentage using equation (3) iterations can be summarized as shown in table below:

ixf(x)f′(x)f″(x)εa0−1−29−161−0.43750.78711.865−13.04616.162−0.34650.8408−0.0931−13.2850.113−0.34720.8408−0.0099−13.28000

Thus, within four iterations, the result converges to true value −0.3472.

Therefore, the maximum of the function f(x)=−x4−2x3−8x2−5x is −0.3472.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 4

Textbook Question

Chapter 13, Problem 8P

Employ the following methods to find the maximum of the function from Prob. 13.7:

(a) Golden-section search ( x l = − 2 , x u = 1 , ε s = 1 % ) .

(b) Parabolic interpolation ( x 0 = − 2 , x 1 = − 1 , x 2 = 1 , iterations = 4 ) . Select new points sequentially as in the secant method.

(c) Newton's method ( x 0 = − 1 , ε s = 1 % )

(a)

Expert Solution

To determine

To calculate: The maximum of the function f(x)=−x4−2x3−8x2−5x with the help of Golden-section search. Take initial guesses as xl=−2,xu=1, andan acceptable error as et=1%.

Answer to Problem 8P

Solution:

The maximum of the function f(x)=−x4−2x3−8x2−5x is, x=−0.3468 after 14 iterations.

Explanation of Solution

Given information:

The function f(x)=−x4−2x3−8x2−5x. Initial guesses xl=−2,xu=1 and value of acceptable error is et=1%.

Formula used:

The golden-search method with two initial guesses, xl and xu that encloses one local extreme of function f(x). Two interior points x1 and x2 with the golden ratio d is given by,

d=5−12( xu−xl)x1=xl+dx2=xu−d

Evaluated the function at the above two interior points. Two results can occur,

If f(x1)>f(x2), then x2 becomes the new xl for the next iteration.

If f(x2)>f(x1), then x1 becomes the new xu for the next iteration.

Calculation:

Consider function f(x), in standard form as

f(x)=−x4−2x3−8x2−5x

With xl=−2 , xu=1 and εs=1%

Iteration 1: First golden ratio is used to create two interior points as,

d=5−12( xu−xl)=5−12( 1−(−2))=3(5−12)=1.8541

The two interior points are as follows:

First point is,

x1=xl+d=−2+1.8541=−0.1459

Second point is,

x2=xu−d=1−1.8541=−0.8541

Now, comparing the value of function at these interior points as shown below:

For x=−0.1459,

f(x1)=−x14−2x13−−8x12−5x1=−(−0.1459)4−2(−0.1459)3−8(−0.1459)2−5(−0.1459)=0.5650

For x=−0.8541,

f(x2)=−x24−2x23−8x22−5x2=−(−0.8541)4−2(−0.8541)3−8(−0.8541)2−5(−0.8541)=−0.8514

As f(x1)>f(x2),

Therefore, the maximum is in the interval defined by x2,x1 and xu

Where, x1 is the optimum.

The error at this point can be computed as follow:

εs=(1−R)| xu−xlxopt |⋅100, where R=5−12εs=(1−5−12)| 1−(−2)−0.1459 |⋅100εs=(1−5−12)(3−0.1459)⋅100εs=785.40%

Therefore, the domain of x to the left of x2 from −2 to x2 can be eliminated; because it does not contain the maximum value.

For this case, x2= −0.8541 becomes new xl for the next round.

Iteration 2: Here, xl=−0.8541 and xu=1. The first golden ratio is used to create two interior points as follow:

d=5−12( xu−xl)=5−12( 1−(−0.8541 ))=1.1459

The two new interior points are as follows:

First point is,

x1=xl+d=−0.8541 +1.1459=0.2918

Second point is,

x2=xu−d=1−1.1459=−0.1459

Now, comparing the value of function at these interior points as shown below:

For x=0.2918,

f(x1)=−x14−2x13−−8x12−5x1=−(0.2918)4−2(0.2918)3−8(0.2918)2−5(0.2918)=−2.197

For x=0.1459,

f(x2)=−x24−2x23−8x22−5x2=−(−0.1459)4−2(−0.1459)3−8(−0.1459)2−5(−0.1459)=0.5650

As f(x2)>f(x1)

For this case, x1=0.2918 become new xu, for the next round.

Now,

εs≃(1−R)|xu−xlxopt|⋅100,where R=5−12εs≃(1−5−12)|1−(−0.8541)−0.1459|⋅100εs≃(1−5−12)(1.8541−0.1459)⋅100εs≃485.40%

Iteration 3: Here, xl=−0.8541 and xu=0.2918. The first golden ratio is used to create two interior points as follow:

d=5−12( xu−xl)=5−12( 0.2918−(−0.8541))=0.7082

The two new interior points are as follows:

First interior point is,

x1=xl+d=−0.8541+0.7082=−0.1459

Second interior point is,

x2=xu−d=0.2918−0.7082=−0.4164

Now, comparing the value of function at these interior points as shown below:

For x=0.1459,

f(x1)=−x14−2x13−−8x12−5x1=−(−0.1459)4−2(−0.1459)3−8(−0.1459)2−5(−0.1459)=0.5650

For x=0.4164,

f(x2)=−x24−2x23−8x22−5x2=−(−0.4164)4−2(−0.4164)3−8(−0.4164)2−5(−0.4164)=0.8090

As f(x2)>f(x1)

Therefore, for this case, x1=−0.1459 becomes new xu for the next round.

Proceeding like this the iterations can be tabulated below as:

ixl−f(xl)x2f(x2)x1f(x1)xuf(xu)xoptεscomparison1−2−22−0.8541−0.8514−0.14590.56501−16−0.1459785.41f(x1)>f(x2)2−0.8541−0.8514−0.14590.56500.2918−2.1971−16−0.1459485.41f(x2)>f(x1)3−0.8541−0.8514−0.41640.8090−0.14590.56500.2918−2.197−0.4164105.11f(x2)>f(x1)4−0.8541−0.8514−0.58360.4750−0.41640.8090−0.14590.5650−0.416464.96f(x1)>f(x2)5−0.58360.4750−0.41640.8090−0.31310.8330−0.14590.5650−0.313153.40f(x1)>f(x2)6−0.41640.8090−0.31310.8330−0.24920.7760−0.14590.5650−0.313133f(x2)>f(x1)7−0.41640.8090−0.35250.8410−0.31310.8330−0.24920.7764−0.352518.11f(x2)>f(x1)8−0.41640.8090−0.37690.8351−0.35250.8410−0.31310.8330−0.352511.19f(x1)>f(x2)9−0.37690.8351−0.35250.8410−0.33750.8401−0.31310.8330−0.35256.92f(x2)>f(x1)10−0.37690.8351−0.36190.8393−0.35250.8410−0.33750.8401−0.35254.28f(x1)>f(x2)11−0.36190.8393−0.35250.8410−0.34680.8410−0.33750.8401−0.34682.69f(x1)>f(x2)12−0.35250.8410−0.34680.8410−0.34320.8412−0.33750.8401−0.34681.66f(x1)>f(x2)13−0.35250.8410−0.34900.8411−0.34680.8410−0.34320.8412−0.34681.03f(x1)>f(x2)14−0.34900.8411−0.34680.8410−0.34540.8411−0.3432−0.8412−0.34680.63f(x1)>f(x2)

Thus, the result converges to true value x=−0.3468.

(b)

Expert Solution

To determine

To calculate: The maximum of the function f(x)=−x4−2x3−8x2−5x with the help of parabolic interpolation. Take initial guesses as x0=−2,x1=−1,x2=1.

Answer to Problem 8P

Solution:

The maximum of the function f(x)=−x4−2x3−8x2−5x is, x=−0.3472 after 4 iterations.

Explanation of Solution

Given information:

The function f(x)=−x4−2x3−8x2−5x. Initial guesses x0=−2,x1=−1,x2=1 and the number of iterations is, 4.

Formula used:

Consider three points jointly bracket an optimum, thus a unique parabola through these three points can be determined. On differentiating and setting it equal to zero estimate of optimal can be computed.

Consider x0,x1 and x2 as initial guesses and x3 be the value of x that corresponds to the maximum value of quadratic which fit into guesses, then,

x3=f(x0)(x12−x22)+f(x1)(x22−x02)+f(x2)(x02−x12)2f(x0)(x1−x2)+2f(x1)(x2−x0)+2f(x2)(x0−x1) …… (1)

Calculation:

Consider function f(x), in standard form as:

f(x)=−x4−2x3−8x2−5x

With initial guesses x0=−2,x1=−1,x2=1.

Iteration 1: Function values at these three initial points is,

For x=−2,

f(x0)=f(−2)=−(−2)4−2(−2)3−8(−2)2−5(−2)=−22

For x=−1,

f(x1)=f(−1)=−(−1)4−2(−1)3−8(−1)2−5(−1)=−2

For x=1,

f(x2)=f(1)=−(1)4−2(1)3−8(1)2−5(1)=−16

Substitute the value of x0,x1,x2 and f(x0),f(x1),f(x2) in equation (1) to get value of x3,

x3=f(x0)(x12−x22)+f(x1)(x22−x02)+f(x2)(x02−x12)2f(x0)(x1−x2)+2f(x1)(x2−x0)+2f(x2)(x0−x1)=(−22)((−1)2−12)+(−2)(12−(−2)2)+(−16)((−2)2−(−1)2)2(−22)((−1)−1)+2(−2)(1−(−2))+2(−16)((−2)−(−1))=−0.3889

The value of function at x3 is,

f(x3)=f(−0.3889)=−(−0.3889)4−2(−0.3889)3−8(−0.3889)2−5(−0.3889)=0.8293

Therefore, f(x3)>f(x1) then the new value of x is to the right of intermediate point x1 and the lower guess x0 is discarded.

Iteration 2: Now the initial guesses are x0=−1,x1=−0.3889,x2=1

Function values at these three initial points are,

For x=−1,

f(x0)=f(−1)=−(−1)4−2(−1)3−8(−1)2−5(−1)=−2

For x=−0.3889,

f(x1)=f(−0.3889)=−(−0.3889)4−2(−0.3889)3−8(−0.3889)2−5(−0.3889)=0.8293

For x=1,

f(x2)=f(1)=−(1)4−2(1)3−8(1)2−5(1)=−16

Substitute the value of x0,x1,x2 and f(x0),f(x1),f(x2) in equation (1) to get value of x3,

x3=f(x0)(x12−x22)+f(x1)(x22−x02)+f(x2)(x02−x12)2f(x0)(x1−x2)+2f(x1)(x2−x0)+2f(x2)(x0−x1)=(−2)((−0.3889)2−12)+(0.8293)(12−(−1)2)+(−16)((−1)2−(−0.3889)2)2(−2)((−0.3889)−1)+2(0.8293)(1−(−1))+2(−16)((−1)−(−0.3889))=−0.4179

The value of the provided function at x3 is,

f(x3)=f(−0.4179)=−(−0.4179)4−2(−0.4179)3−8(−0.4179)2−5(−0.4179)=0.8077

Therefore, f(x1)>f(x3) then the new value of x is to the right of intermediate point x1 and the upper guess x2 is discarded.

Iteration 3: Now the initial guesses are x0=−1,x1=−0.4179,x2=−0.3889

Function values at these three initial points is,

For x=−1,

f(x0)=f(−1)=−(−1)4−2(−1)3−8(−1)2−5(−1)=−2

The function for x1

f(x1)=f(−0.4179)=−(−0.4179)4−2(−0.4179)3−8(−0.4179)2−5(−0.4179)=0.8077

And for x2,

f(x2)=f(−0.3889)=−(−0.3889)4−2(−0.3889)3−8(−0.3889)2−5(−0.3889)=0.8293

Substitute the value of x0,x1,x2 and f(x0),f(x1),f(x2) in equation (1) to get value of x3,

x3=f(x0)(x12−x22)+f(x1)(x22−x02)+f(x2)(x02−x12)2f(x0)(x1−x2)+2f(x1)(x2−x0)+2f(x2)(x0−x1)={(−2)((−0.4179)2−(−0.3889)2)+(0.8077)((−0.3889)2−(−1)2)+(0.8293)((−1)2−(−0.4179)2)}{2(−2)((−0.4179)−(−0.3889))+2(0.8077)((−0.3889)−(−1))+2(0.8293)((−1)−(−0.4179))}=−0.3476

And value of function at x3 is,

f(x3)=f(−0.3476)=−(−0.3476)4−2(−0.3476)3−8(−0.3476)2−5(−0.3476)=0.8408

Therefore, f(x3)>f(x1) then the new value of x is to the right of intermediate point x1 and the lower guess x0 is discarded.

Iteration 4: Now the initial guesses are x0=−0.4179,x1=−0.3889,x2=−0.3476

Function values at these three initial points is,

For x=−0.4179,

f(x0)=f(−0.4179)=−(−0.4179)4−2(−0.4179)3−8(−0.4179)2−5(−0.4179)=0.8077

For x=−0.3889,

f(x1)=f(−0.3889)=−(−0.3889)4−2(−0.3889)3−8(−0.3889)2−5(−0.3889)=0.8293

For x=−0.3476,

f(x2)=f(−0.3476)=−(−0.3476)4−2(−0.3476)3−8(−0.3476)2−5(−0.3476)=0.8408

Substitute the value of x0,x1,x2 and f(x0),f(x1),f(x2) in equation (1) to get value of x3,

x3=(f(x0)(x12−x22)+f(x1)(x22−x02)+f(x2)(x02−x12))(2f(x0)(x1−x2)+2f(x1)(x2−x0)+2f(x2)(x0−x1))=((0.8077)((−0.3889)2−(−0.3476)2)+(0.8293)((−0.3476)2−(−0.4179)2)+(0.8408)((−0.4179)2−(−0.3889)2))(2(0.8077)((−0.3889)−(−0.3476))+2(0.8293)((−0.3476)−(−0.4179))+2(0.8408)((−0.4179)−(−0.3889)))=−0.3472

And value of function at x3 is,

f(x3)=f(−0.3472)=−(−0.3472)4−2(−0.3472)3−8(−0.3472)2−5(−0.3472)=0.8408

Therefore, f(x1)>f(x3) then the new value of x is to the right of intermediate point x1 and the lower guess x0 is discarded.

And the process continues with a summary shown below in a table:

ix0f(x0)x1f(x1)x2f(x2)x3f(x3)εa(%)1−2−22−1−21−16−0.38890.82932−1−2−0.38890.82931−16−0.41790.80776.943−1−2−0.41790.8077−0.38890.8293−0.34760.840820.224−0.41790.8077−0.38890.8293−0.34760.8408−0.34720.84080.115

Thus, after four iterations result is converging to true value x=−0.3472.

(c)

Expert Solution

To determine

To calculate: The maximum of the function f(x)=−x4−2x3−8x2−5x with the help of Newton’s method. Take initial guess as x0=−1.

Answer to Problem 8P

Solution:

The maximum of the function f(x)=−x4−2x3−8x2−5x is, −0.3472.

Explanation of Solution

Given information:

The function f(x)=−x4−2x3−8x2−5x. Initial guess is x0=−1 and the acceptable error is, et=1%.

Formula used:

Newton Method is similar to Newton Raphson as it does not require initial guesses that bracket the optimum solution.

For any function f(x) optimal solution can be attained by performing iterations following the formula given below:

xi+1=xi−f′(xi)f″(xi) ...... (1)

Calculation:

Consider function f(x),

f(x)=−x4−2x3−8x2−5x

With initial guesses x0=−1,εt=1%.

First and second derivatives of function that is, f′(x) and f″(x), can be evaluated as:

f′(x)= −4x3−6x2−16x−5f″(x)=−12x2−12x−16

Iteration 1:Initially for x0=−1, values of first and second derivatives of function are as follows:

f′(x)= −4x3−6x2−16x−5f′(−1)=−4(−1)3−6(−1)2−16(−1)−5=9

For second derivative,

f″(x)=−12x2−12x−16f″(−1)=−12(−1)2−12(−1)−16=−16

Therefore, x1 is calculated using formula given as equation (2),

x1=x0−f′(x0)f″(x0)=−1−(9)(−16)=−0.4375

And

f(x1)=−x14−2x13−8x12−5x1=−(−0.4375)4−2(−0.4375)3−8(−0.4375)2−5(−0.4375)=0.7871

Iteration 2:Now for x1=−0.4375, values of first and second derivatives of function are as follows:

f′(x1)= −4x13−6x12−16x1−5f′(−0.4375)=−4(−0.4375)3−6(−0.4375)2−16(−0.4375)−5=1.1865

For second derivative,

f″(x)=−12x2−12x−16f″(−0.4375)=−12(−0.4375)2−12(−0.4375)−16=−13.045

Therefore, x2 is calculated using formula given as equation (2)

x2=x1−f′(x1)f″(x1)=−0.4375−(1.1865)(−13.045)=−0.3465

And

f(x2)=−x24−2x23−8x22−5x2=−(−0.3465)4−2(−0.3465)3−8(−0.3465)2−5(−0.3465)=0.8408

Iteration 3:Now for x2=−0.3465, values of first and second derivatives of function are as follows:

f′(x2)= −4x23−6x22−16x2−5f′(−0.3465)=−4(−0.3465)3−6(−0.3465)2−16(−0.3465)−5=−0.0099

For second derivative,

f″(x2)=−12x22−12x2−16f″(−0.3465)=−12(−0.3465)2−12(−0.3465)−16=−13.2827

Therefore, x3 is calculated using formula given as equation (2)

x3=x2−f′(x2)f″(x2)=−0.3465−(−0.0099)(−13.2827)=−0.3472

And

f(x3)=−x34−2x33−8x32−5x3=−(−0.3472)4−2(−0.3472)3−8(−0.3472)2−5(−0.3472)=0.8408

Iteration 4:Now for x3=−0.3472, values of first and second derivatives of function are as follows:

f′(x3)= −4x33−6x32−16x3−5f′(−0.3472)=−4(−0.3472)3−6(−0.3472)2−16(−0.3472)−5=−0.00067

For second derivative,

f″(x2)=−12x22−12x2−16f″(−0.3472)=−12(−0.3472)2−12(−0.3472)−16=−13.2801

Therefore, x4 is calculated using formula given as equation (2)

x4=x3−f′(x3)f″(x3)=−0.3472−(−0.00067)(−13.2801)=−0.347250

Maintaining the error percentage using equation (3) iterations can be summarized as shown in table below:

ixf(x)f′(x)f″(x)εa0−1−29−161−0.43750.78711.865−13.04616.162−0.34650.8408−0.0931−13.2850.113−0.34720.8408−0.0099−13.28000

Thus, within four iterations, the result converges to true value −0.3472.

Therefore, the maximum of the function f(x)=−x4−2x3−8x2−5x is −0.3472.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

(a)

Expert Solution

To determine

To calculate: The maximum of the function f(x)=−x4−2x3−8x2−5x with the help of Golden-section search. Take initial guesses as xl=−2,xu=1, andan acceptable error as et=1%.

Answer to Problem 8P

Solution:

The maximum of the function f(x)=−x4−2x3−8x2−5x is, x=−0.3468 after 14 iterations.

Explanation of Solution

Given information:

The function f(x)=−x4−2x3−8x2−5x. Initial guesses xl=−2,xu=1 and value of acceptable error is et=1%.

Formula used:

The golden-search method with two initial guesses, xl and xu that encloses one local extreme of function f(x). Two interior points x1 and x2 with the golden ratio d is given by,

d=5−12( xu−xl)x1=xl+dx2=xu−d

Evaluated the function at the above two interior points. Two results can occur,

If f(x1)>f(x2), then x2 becomes the new xl for the next iteration.

If f(x2)>f(x1), then x1 becomes the new xu for the next iteration.

Calculation:

Consider function f(x), in standard form as

f(x)=−x4−2x3−8x2−5x

With xl=−2 , xu=1 and εs=1%

Iteration 1: First golden ratio is used to create two interior points as,

d=5−12( xu−xl)=5−12( 1−(−2))=3(5−12)=1.8541

The two interior points are as follows:

First point is,

x1=xl+d=−2+1.8541=−0.1459

Second point is,

x2=xu−d=1−1.8541=−0.8541

Now, comparing the value of function at these interior points as shown below:

For x=−0.1459,

f(x1)=−x14−2x13−−8x12−5x1=−(−0.1459)4−2(−0.1459)3−8(−0.1459)2−5(−0.1459)=0.5650

For x=−0.8541,

f(x2)=−x24−2x23−8x22−5x2=−(−0.8541)4−2(−0.8541)3−8(−0.8541)2−5(−0.8541)=−0.8514

As f(x1)>f(x2),

Therefore, the maximum is in the interval defined by x2,x1 and xu

Where, x1 is the optimum.

The error at this point can be computed as follow:

εs=(1−R)| xu−xlxopt |⋅100, where R=5−12εs=(1−5−12)| 1−(−2)−0.1459 |⋅100εs=(1−5−12)(3−0.1459)⋅100εs=785.40%

Therefore, the domain of x to the left of x2 from −2 to x2 can be eliminated; because it does not contain the maximum value.

For this case, x2= −0.8541 becomes new xl for the next round.

Iteration 2: Here, xl=−0.8541 and xu=1. The first golden ratio is used to create two interior points as follow:

d=5−12( xu−xl)=5−12( 1−(−0.8541 ))=1.1459

The two new interior points are as follows:

First point is,

x1=xl+d=−0.8541 +1.1459=0.2918

Second point is,

x2=xu−d=1−1.1459=−0.1459

Now, comparing the value of function at these interior points as shown below:

For x=0.2918,

f(x1)=−x14−2x13−−8x12−5x1=−(0.2918)4−2(0.2918)3−8(0.2918)2−5(0.2918)=−2.197

For x=0.1459,

f(x2)=−x24−2x23−8x22−5x2=−(−0.1459)4−2(−0.1459)3−8(−0.1459)2−5(−0.1459)=0.5650

As f(x2)>f(x1)

For this case, x1=0.2918 become new xu, for the next round.

Now,

εs≃(1−R)|xu−xlxopt|⋅100,where R=5−12εs≃(1−5−12)|1−(−0.8541)−0.1459|⋅100εs≃(1−5−12)(1.8541−0.1459)⋅100εs≃485.40%

Iteration 3: Here, xl=−0.8541 and xu=0.2918. The first golden ratio is used to create two interior points as follow:

d=5−12( xu−xl)=5−12( 0.2918−(−0.8541))=0.7082

The two new interior points are as follows:

First interior point is,

x1=xl+d=−0.8541+0.7082=−0.1459

Second interior point is,

x2=xu−d=0.2918−0.7082=−0.4164

Now, comparing the value of function at these interior points as shown below:

For x=0.1459,

f(x1)=−x14−2x13−−8x12−5x1=−(−0.1459)4−2(−0.1459)3−8(−0.1459)2−5(−0.1459)=0.5650

For x=0.4164,

f(x2)=−x24−2x23−8x22−5x2=−(−0.4164)4−2(−0.4164)3−8(−0.4164)2−5(−0.4164)=0.8090

As f(x2)>f(x1)

Therefore, for this case, x1=−0.1459 becomes new xu for the next round.

Proceeding like this the iterations can be tabulated below as:

ixl−f(xl)x2f(x2)x1f(x1)xuf(xu)xoptεscomparison1−2−22−0.8541−0.8514−0.14590.56501−16−0.1459785.41f(x1)>f(x2)2−0.8541−0.8514−0.14590.56500.2918−2.1971−16−0.1459485.41f(x2)>f(x1)3−0.8541−0.8514−0.41640.8090−0.14590.56500.2918−2.197−0.4164105.11f(x2)>f(x1)4−0.8541−0.8514−0.58360.4750−0.41640.8090−0.14590.5650−0.416464.96f(x1)>f(x2)5−0.58360.4750−0.41640.8090−0.31310.8330−0.14590.5650−0.313153.40f(x1)>f(x2)6−0.41640.8090−0.31310.8330−0.24920.7760−0.14590.5650−0.313133f(x2)>f(x1)7−0.41640.8090−0.35250.8410−0.31310.8330−0.24920.7764−0.352518.11f(x2)>f(x1)8−0.41640.8090−0.37690.8351−0.35250.8410−0.31310.8330−0.352511.19f(x1)>f(x2)9−0.37690.8351−0.35250.8410−0.33750.8401−0.31310.8330−0.35256.92f(x2)>f(x1)10−0.37690.8351−0.36190.8393−0.35250.8410−0.33750.8401−0.35254.28f(x1)>f(x2)11−0.36190.8393−0.35250.8410−0.34680.8410−0.33750.8401−0.34682.69f(x1)>f(x2)12−0.35250.8410−0.34680.8410−0.34320.8412−0.33750.8401−0.34681.66f(x1)>f(x2)13−0.35250.8410−0.34900.8411−0.34680.8410−0.34320.8412−0.34681.03f(x1)>f(x2)14−0.34900.8411−0.34680.8410−0.34540.8411−0.3432−0.8412−0.34680.63f(x1)>f(x2)

Thus, the result converges to true value x=−0.3468.

(b)

Expert Solution

To determine

To calculate: The maximum of the function f(x)=−x4−2x3−8x2−5x with the help of parabolic interpolation. Take initial guesses as x0=−2,x1=−1,x2=1.

Answer to Problem 8P

Solution:

The maximum of the function f(x)=−x4−2x3−8x2−5x is, x=−0.3472 after 4 iterations.

Explanation of Solution

Given information:

The function f(x)=−x4−2x3−8x2−5x. Initial guesses x0=−2,x1=−1,x2=1 and the number of iterations is, 4.

Formula used:

Consider three points jointly bracket an optimum, thus a unique parabola through these three points can be determined. On differentiating and setting it equal to zero estimate of optimal can be computed.

Consider x0,x1 and x2 as initial guesses and x3 be the value of x that corresponds to the maximum value of quadratic which fit into guesses, then,

x3=f(x0)(x12−x22)+f(x1)(x22−x02)+f(x2)(x02−x12)2f(x0)(x1−x2)+2f(x1)(x2−x0)+2f(x2)(x0−x1) …… (1)

Calculation:

Consider function f(x), in standard form as:

f(x)=−x4−2x3−8x2−5x

With initial guesses x0=−2,x1=−1,x2=1.

Iteration 1: Function values at these three initial points is,

For x=−2,

f(x0)=f(−2)=−(−2)4−2(−2)3−8(−2)2−5(−2)=−22

For x=−1,

f(x1)=f(−1)=−(−1)4−2(−1)3−8(−1)2−5(−1)=−2

For x=1,

f(x2)=f(1)=−(1)4−2(1)3−8(1)2−5(1)=−16

Substitute the value of x0,x1,x2 and f(x0),f(x1),f(x2) in equation (1) to get value of x3,

x3=f(x0)(x12−x22)+f(x1)(x22−x02)+f(x2)(x02−x12)2f(x0)(x1−x2)+2f(x1)(x2−x0)+2f(x2)(x0−x1)=(−22)((−1)2−12)+(−2)(12−(−2)2)+(−16)((−2)2−(−1)2)2(−22)((−1)−1)+2(−2)(1−(−2))+2(−16)((−2)−(−1))=−0.3889

The value of function at x3 is,

f(x3)=f(−0.3889)=−(−0.3889)4−2(−0.3889)3−8(−0.3889)2−5(−0.3889)=0.8293

Therefore, f(x3)>f(x1) then the new value of x is to the right of intermediate point x1 and the lower guess x0 is discarded.

Iteration 2: Now the initial guesses are x0=−1,x1=−0.3889,x2=1

Function values at these three initial points are,

For x=−1,

f(x0)=f(−1)=−(−1)4−2(−1)3−8(−1)2−5(−1)=−2

For x=−0.3889,

f(x1)=f(−0.3889)=−(−0.3889)4−2(−0.3889)3−8(−0.3889)2−5(−0.3889)=0.8293

For x=1,

f(x2)=f(1)=−(1)4−2(1)3−8(1)2−5(1)=−16

Substitute the value of x0,x1,x2 and f(x0),f(x1),f(x2) in equation (1) to get value of x3,

x3=f(x0)(x12−x22)+f(x1)(x22−x02)+f(x2)(x02−x12)2f(x0)(x1−x2)+2f(x1)(x2−x0)+2f(x2)(x0−x1)=(−2)((−0.3889)2−12)+(0.8293)(12−(−1)2)+(−16)((−1)2−(−0.3889)2)2(−2)((−0.3889)−1)+2(0.8293)(1−(−1))+2(−16)((−1)−(−0.3889))=−0.4179

The value of the provided function at x3 is,

f(x3)=f(−0.4179)=−(−0.4179)4−2(−0.4179)3−8(−0.4179)2−5(−0.4179)=0.8077

Therefore, f(x1)>f(x3) then the new value of x is to the right of intermediate point x1 and the upper guess x2 is discarded.

Iteration 3: Now the initial guesses are x0=−1,x1=−0.4179,x2=−0.3889

Function values at these three initial points is,

For x=−1,

f(x0)=f(−1)=−(−1)4−2(−1)3−8(−1)2−5(−1)=−2

The function for x1

f(x1)=f(−0.4179)=−(−0.4179)4−2(−0.4179)3−8(−0.4179)2−5(−0.4179)=0.8077

And for x2,

f(x2)=f(−0.3889)=−(−0.3889)4−2(−0.3889)3−8(−0.3889)2−5(−0.3889)=0.8293

Substitute the value of x0,x1,x2 and f(x0),f(x1),f(x2) in equation (1) to get value of x3,

x3=f(x0)(x12−x22)+f(x1)(x22−x02)+f(x2)(x02−x12)2f(x0)(x1−x2)+2f(x1)(x2−x0)+2f(x2)(x0−x1)={(−2)((−0.4179)2−(−0.3889)2)+(0.8077)((−0.3889)2−(−1)2)+(0.8293)((−1)2−(−0.4179)2)}{2(−2)((−0.4179)−(−0.3889))+2(0.8077)((−0.3889)−(−1))+2(0.8293)((−1)−(−0.4179))}=−0.3476

And value of function at x3 is,

f(x3)=f(−0.3476)=−(−0.3476)4−2(−0.3476)3−8(−0.3476)2−5(−0.3476)=0.8408

Therefore, f(x3)>f(x1) then the new value of x is to the right of intermediate point x1 and the lower guess x0 is discarded.

Iteration 4: Now the initial guesses are x0=−0.4179,x1=−0.3889,x2=−0.3476

Function values at these three initial points is,

For x=−0.4179,

f(x0)=f(−0.4179)=−(−0.4179)4−2(−0.4179)3−8(−0.4179)2−5(−0.4179)=0.8077

For x=−0.3889,

f(x1)=f(−0.3889)=−(−0.3889)4−2(−0.3889)3−8(−0.3889)2−5(−0.3889)=0.8293

For x=−0.3476,

f(x2)=f(−0.3476)=−(−0.3476)4−2(−0.3476)3−8(−0.3476)2−5(−0.3476)=0.8408

Substitute the value of x0,x1,x2 and f(x0),f(x1),f(x2) in equation (1) to get value of x3,

x3=(f(x0)(x12−x22)+f(x1)(x22−x02)+f(x2)(x02−x12))(2f(x0)(x1−x2)+2f(x1)(x2−x0)+2f(x2)(x0−x1))=((0.8077)((−0.3889)2−(−0.3476)2)+(0.8293)((−0.3476)2−(−0.4179)2)+(0.8408)((−0.4179)2−(−0.3889)2))(2(0.8077)((−0.3889)−(−0.3476))+2(0.8293)((−0.3476)−(−0.4179))+2(0.8408)((−0.4179)−(−0.3889)))=−0.3472

And value of function at x3 is,

f(x3)=f(−0.3472)=−(−0.3472)4−2(−0.3472)3−8(−0.3472)2−5(−0.3472)=0.8408

Therefore, f(x1)>f(x3) then the new value of x is to the right of intermediate point x1 and the lower guess x0 is discarded.

And the process continues with a summary shown below in a table:

ix0f(x0)x1f(x1)x2f(x2)x3f(x3)εa(%)1−2−22−1−21−16−0.38890.82932−1−2−0.38890.82931−16−0.41790.80776.943−1−2−0.41790.8077−0.38890.8293−0.34760.840820.224−0.41790.8077−0.38890.8293−0.34760.8408−0.34720.84080.115

Thus, after four iterations result is converging to true value x=−0.3472.

(c)

Expert Solution

To determine

To calculate: The maximum of the function f(x)=−x4−2x3−8x2−5x with the help of Newton’s method. Take initial guess as x0=−1.

Answer to Problem 8P

Solution:

The maximum of the function f(x)=−x4−2x3−8x2−5x is, −0.3472.

Explanation of Solution

Given information:

The function f(x)=−x4−2x3−8x2−5x. Initial guess is x0=−1 and the acceptable error is, et=1%.

Formula used:

Newton Method is similar to Newton Raphson as it does not require initial guesses that bracket the optimum solution.

For any function f(x) optimal solution can be attained by performing iterations following the formula given below:

xi+1=xi−f′(xi)f″(xi) ...... (1)

Calculation:

Consider function f(x),

f(x)=−x4−2x3−8x2−5x

With initial guesses x0=−1,εt=1%.

First and second derivatives of function that is, f′(x) and f″(x), can be evaluated as:

f′(x)= −4x3−6x2−16x−5f″(x)=−12x2−12x−16

Iteration 1:Initially for x0=−1, values of first and second derivatives of function are as follows:

f′(x)= −4x3−6x2−16x−5f′(−1)=−4(−1)3−6(−1)2−16(−1)−5=9

For second derivative,

f″(x)=−12x2−12x−16f″(−1)=−12(−1)2−12(−1)−16=−16

Therefore, x1 is calculated using formula given as equation (2),

x1=x0−f′(x0)f″(x0)=−1−(9)(−16)=−0.4375

And

f(x1)=−x14−2x13−8x12−5x1=−(−0.4375)4−2(−0.4375)3−8(−0.4375)2−5(−0.4375)=0.7871

Iteration 2:Now for x1=−0.4375, values of first and second derivatives of function are as follows:

f′(x1)= −4x13−6x12−16x1−5f′(−0.4375)=−4(−0.4375)3−6(−0.4375)2−16(−0.4375)−5=1.1865

For second derivative,

f″(x)=−12x2−12x−16f″(−0.4375)=−12(−0.4375)2−12(−0.4375)−16=−13.045

Therefore, x2 is calculated using formula given as equation (2)

x2=x1−f′(x1)f″(x1)=−0.4375−(1.1865)(−13.045)=−0.3465

And

f(x2)=−x24−2x23−8x22−5x2=−(−0.3465)4−2(−0.3465)3−8(−0.3465)2−5(−0.3465)=0.8408

Iteration 3:Now for x2=−0.3465, values of first and second derivatives of function are as follows:

f′(x2)= −4x23−6x22−16x2−5f′(−0.3465)=−4(−0.3465)3−6(−0.3465)2−16(−0.3465)−5=−0.0099

For second derivative,

f″(x2)=−12x22−12x2−16f″(−0.3465)=−12(−0.3465)2−12(−0.3465)−16=−13.2827

Therefore, x3 is calculated using formula given as equation (2)

x3=x2−f′(x2)f″(x2)=−0.3465−(−0.0099)(−13.2827)=−0.3472

And

f(x3)=−x34−2x33−8x32−5x3=−(−0.3472)4−2(−0.3472)3−8(−0.3472)2−5(−0.3472)=0.8408

Iteration 4:Now for x3=−0.3472, values of first and second derivatives of function are as follows:

f′(x3)= −4x33−6x32−16x3−5f′(−0.3472)=−4(−0.3472)3−6(−0.3472)2−16(−0.3472)−5=−0.00067

For second derivative,

f″(x2)=−12x22−12x2−16f″(−0.3472)=−12(−0.3472)2−12(−0.3472)−16=−13.2801

Therefore, x4 is calculated using formula given as equation (2)

x4=x3−f′(x3)f″(x3)=−0.3472−(−0.00067)(−13.2801)=−0.347250

Maintaining the error percentage using equation (3) iterations can be summarized as shown in table below:

ixf(x)f′(x)f″(x)εa0−1−29−161−0.43750.78711.865−13.04616.162−0.34650.8408−0.0931−13.2850.113−0.34720.8408−0.0099−13.28000

Thus, within four iterations, the result converges to true value −0.3472.

Therefore, the maximum of the function f(x)=−x4−2x3−8x2−5x is −0.3472.

Answer 8

(a)

Expert Solution

To determine

To calculate: The maximum of the function f(x)=−x4−2x3−8x2−5x with the help of Golden-section search. Take initial guesses as xl=−2,xu=1, andan acceptable error as et=1%.

Answer to Problem 8P

Solution:

The maximum of the function f(x)=−x4−2x3−8x2−5x is, x=−0.3468 after 14 iterations.

Explanation of Solution

Given information:

The function f(x)=−x4−2x3−8x2−5x. Initial guesses xl=−2,xu=1 and value of acceptable error is et=1%.

Formula used:

The golden-search method with two initial guesses, xl and xu that encloses one local extreme of function f(x). Two interior points x1 and x2 with the golden ratio d is given by,

d=5−12( xu−xl)x1=xl+dx2=xu−d

Evaluated the function at the above two interior points. Two results can occur,

If f(x1)>f(x2), then x2 becomes the new xl for the next iteration.

If f(x2)>f(x1), then x1 becomes the new xu for the next iteration.

Calculation:

Consider function f(x), in standard form as

f(x)=−x4−2x3−8x2−5x

With xl=−2 , xu=1 and εs=1%

Iteration 1: First golden ratio is used to create two interior points as,

d=5−12( xu−xl)=5−12( 1−(−2))=3(5−12)=1.8541

The two interior points are as follows:

First point is,

x1=xl+d=−2+1.8541=−0.1459

Second point is,

x2=xu−d=1−1.8541=−0.8541

Now, comparing the value of function at these interior points as shown below:

For x=−0.1459,

f(x1)=−x14−2x13−−8x12−5x1=−(−0.1459)4−2(−0.1459)3−8(−0.1459)2−5(−0.1459)=0.5650

For x=−0.8541,

f(x2)=−x24−2x23−8x22−5x2=−(−0.8541)4−2(−0.8541)3−8(−0.8541)2−5(−0.8541)=−0.8514

As f(x1)>f(x2),

Therefore, the maximum is in the interval defined by x2,x1 and xu

Where, x1 is the optimum.

The error at this point can be computed as follow:

εs=(1−R)| xu−xlxopt |⋅100, where R=5−12εs=(1−5−12)| 1−(−2)−0.1459 |⋅100εs=(1−5−12)(3−0.1459)⋅100εs=785.40%

Therefore, the domain of x to the left of x2 from −2 to x2 can be eliminated; because it does not contain the maximum value.

For this case, x2= −0.8541 becomes new xl for the next round.

Iteration 2: Here, xl=−0.8541 and xu=1. The first golden ratio is used to create two interior points as follow:

d=5−12( xu−xl)=5−12( 1−(−0.8541 ))=1.1459

The two new interior points are as follows:

First point is,

x1=xl+d=−0.8541 +1.1459=0.2918

Second point is,

x2=xu−d=1−1.1459=−0.1459

Now, comparing the value of function at these interior points as shown below:

For x=0.2918,

f(x1)=−x14−2x13−−8x12−5x1=−(0.2918)4−2(0.2918)3−8(0.2918)2−5(0.2918)=−2.197

For x=0.1459,

f(x2)=−x24−2x23−8x22−5x2=−(−0.1459)4−2(−0.1459)3−8(−0.1459)2−5(−0.1459)=0.5650

As f(x2)>f(x1)

For this case, x1=0.2918 become new xu, for the next round.

Now,

εs≃(1−R)|xu−xlxopt|⋅100,where R=5−12εs≃(1−5−12)|1−(−0.8541)−0.1459|⋅100εs≃(1−5−12)(1.8541−0.1459)⋅100εs≃485.40%

Iteration 3: Here, xl=−0.8541 and xu=0.2918. The first golden ratio is used to create two interior points as follow:

d=5−12( xu−xl)=5−12( 0.2918−(−0.8541))=0.7082

The two new interior points are as follows:

First interior point is,

x1=xl+d=−0.8541+0.7082=−0.1459

Second interior point is,

x2=xu−d=0.2918−0.7082=−0.4164

Now, comparing the value of function at these interior points as shown below:

For x=0.1459,

f(x1)=−x14−2x13−−8x12−5x1=−(−0.1459)4−2(−0.1459)3−8(−0.1459)2−5(−0.1459)=0.5650

For x=0.4164,

f(x2)=−x24−2x23−8x22−5x2=−(−0.4164)4−2(−0.4164)3−8(−0.4164)2−5(−0.4164)=0.8090

As f(x2)>f(x1)

Therefore, for this case, x1=−0.1459 becomes new xu for the next round.

Proceeding like this the iterations can be tabulated below as:

ixl−f(xl)x2f(x2)x1f(x1)xuf(xu)xoptεscomparison1−2−22−0.8541−0.8514−0.14590.56501−16−0.1459785.41f(x1)>f(x2)2−0.8541−0.8514−0.14590.56500.2918−2.1971−16−0.1459485.41f(x2)>f(x1)3−0.8541−0.8514−0.41640.8090−0.14590.56500.2918−2.197−0.4164105.11f(x2)>f(x1)4−0.8541−0.8514−0.58360.4750−0.41640.8090−0.14590.5650−0.416464.96f(x1)>f(x2)5−0.58360.4750−0.41640.8090−0.31310.8330−0.14590.5650−0.313153.40f(x1)>f(x2)6−0.41640.8090−0.31310.8330−0.24920.7760−0.14590.5650−0.313133f(x2)>f(x1)7−0.41640.8090−0.35250.8410−0.31310.8330−0.24920.7764−0.352518.11f(x2)>f(x1)8−0.41640.8090−0.37690.8351−0.35250.8410−0.31310.8330−0.352511.19f(x1)>f(x2)9−0.37690.8351−0.35250.8410−0.33750.8401−0.31310.8330−0.35256.92f(x2)>f(x1)10−0.37690.8351−0.36190.8393−0.35250.8410−0.33750.8401−0.35254.28f(x1)>f(x2)11−0.36190.8393−0.35250.8410−0.34680.8410−0.33750.8401−0.34682.69f(x1)>f(x2)12−0.35250.8410−0.34680.8410−0.34320.8412−0.33750.8401−0.34681.66f(x1)>f(x2)13−0.35250.8410−0.34900.8411−0.34680.8410−0.34320.8412−0.34681.03f(x1)>f(x2)14−0.34900.8411−0.34680.8410−0.34540.8411−0.3432−0.8412−0.34680.63f(x1)>f(x2)

Thus, the result converges to true value x=−0.3468.

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

To determine

To calculate: The maximum of the function f(x)=−x4−2x3−8x2−5x with the help of Golden-section search. Take initial guesses as xl=−2,xu=1, andan acceptable error as et=1%.

Answer 13

Answer to Problem 8P

Solution:

The maximum of the function f(x)=−x4−2x3−8x2−5x is, x=−0.3468 after 14 iterations.

Answer 14

Explanation of Solution

Given information:

The function f(x)=−x4−2x3−8x2−5x. Initial guesses xl=−2,xu=1 and value of acceptable error is et=1%.

Formula used:

The golden-search method with two initial guesses, xl and xu that encloses one local extreme of function f(x). Two interior points x1 and x2 with the golden ratio d is given by,

d=5−12( xu−xl)x1=xl+dx2=xu−d

Evaluated the function at the above two interior points. Two results can occur,

If f(x1)>f(x2), then x2 becomes the new xl for the next iteration.

If f(x2)>f(x1), then x1 becomes the new xu for the next iteration.

Calculation:

Consider function f(x), in standard form as

f(x)=−x4−2x3−8x2−5x

With xl=−2 , xu=1 and εs=1%

Iteration 1: First golden ratio is used to create two interior points as,

d=5−12( xu−xl)=5−12( 1−(−2))=3(5−12)=1.8541

The two interior points are as follows:

First point is,

x1=xl+d=−2+1.8541=−0.1459

Second point is,

x2=xu−d=1−1.8541=−0.8541

Now, comparing the value of function at these interior points as shown below:

For x=−0.1459,

f(x1)=−x14−2x13−−8x12−5x1=−(−0.1459)4−2(−0.1459)3−8(−0.1459)2−5(−0.1459)=0.5650

For x=−0.8541,

f(x2)=−x24−2x23−8x22−5x2=−(−0.8541)4−2(−0.8541)3−8(−0.8541)2−5(−0.8541)=−0.8514

As f(x1)>f(x2),

Therefore, the maximum is in the interval defined by x2,x1 and xu

Where, x1 is the optimum.

The error at this point can be computed as follow:

εs=(1−R)| xu−xlxopt |⋅100, where R=5−12εs=(1−5−12)| 1−(−2)−0.1459 |⋅100εs=(1−5−12)(3−0.1459)⋅100εs=785.40%

Therefore, the domain of x to the left of x2 from −2 to x2 can be eliminated; because it does not contain the maximum value.

For this case, x2= −0.8541 becomes new xl for the next round.

Iteration 2: Here, xl=−0.8541 and xu=1. The first golden ratio is used to create two interior points as follow:

d=5−12( xu−xl)=5−12( 1−(−0.8541 ))=1.1459

The two new interior points are as follows:

First point is,

x1=xl+d=−0.8541 +1.1459=0.2918

Second point is,

x2=xu−d=1−1.1459=−0.1459

Now, comparing the value of function at these interior points as shown below:

For x=0.2918,

f(x1)=−x14−2x13−−8x12−5x1=−(0.2918)4−2(0.2918)3−8(0.2918)2−5(0.2918)=−2.197

For x=0.1459,

f(x2)=−x24−2x23−8x22−5x2=−(−0.1459)4−2(−0.1459)3−8(−0.1459)2−5(−0.1459)=0.5650

As f(x2)>f(x1)

For this case, x1=0.2918 become new xu, for the next round.

Now,

εs≃(1−R)|xu−xlxopt|⋅100,where R=5−12εs≃(1−5−12)|1−(−0.8541)−0.1459|⋅100εs≃(1−5−12)(1.8541−0.1459)⋅100εs≃485.40%

Iteration 3: Here, xl=−0.8541 and xu=0.2918. The first golden ratio is used to create two interior points as follow:

d=5−12( xu−xl)=5−12( 0.2918−(−0.8541))=0.7082

The two new interior points are as follows:

First interior point is,

x1=xl+d=−0.8541+0.7082=−0.1459

Second interior point is,

x2=xu−d=0.2918−0.7082=−0.4164

Now, comparing the value of function at these interior points as shown below:

For x=0.1459,

f(x1)=−x14−2x13−−8x12−5x1=−(−0.1459)4−2(−0.1459)3−8(−0.1459)2−5(−0.1459)=0.5650

For x=0.4164,

f(x2)=−x24−2x23−8x22−5x2=−(−0.4164)4−2(−0.4164)3−8(−0.4164)2−5(−0.4164)=0.8090

As f(x2)>f(x1)

Therefore, for this case, x1=−0.1459 becomes new xu for the next round.

Proceeding like this the iterations can be tabulated below as:

ixl−f(xl)x2f(x2)x1f(x1)xuf(xu)xoptεscomparison1−2−22−0.8541−0.8514−0.14590.56501−16−0.1459785.41f(x1)>f(x2)2−0.8541−0.8514−0.14590.56500.2918−2.1971−16−0.1459485.41f(x2)>f(x1)3−0.8541−0.8514−0.41640.8090−0.14590.56500.2918−2.197−0.4164105.11f(x2)>f(x1)4−0.8541−0.8514−0.58360.4750−0.41640.8090−0.14590.5650−0.416464.96f(x1)>f(x2)5−0.58360.4750−0.41640.8090−0.31310.8330−0.14590.5650−0.313153.40f(x1)>f(x2)6−0.41640.8090−0.31310.8330−0.24920.7760−0.14590.5650−0.313133f(x2)>f(x1)7−0.41640.8090−0.35250.8410−0.31310.8330−0.24920.7764−0.352518.11f(x2)>f(x1)8−0.41640.8090−0.37690.8351−0.35250.8410−0.31310.8330−0.352511.19f(x1)>f(x2)9−0.37690.8351−0.35250.8410−0.33750.8401−0.31310.8330−0.35256.92f(x2)>f(x1)10−0.37690.8351−0.36190.8393−0.35250.8410−0.33750.8401−0.35254.28f(x1)>f(x2)11−0.36190.8393−0.35250.8410−0.34680.8410−0.33750.8401−0.34682.69f(x1)>f(x2)12−0.35250.8410−0.34680.8410−0.34320.8412−0.33750.8401−0.34681.66f(x1)>f(x2)13−0.35250.8410−0.34900.8411−0.34680.8410−0.34320.8412−0.34681.03f(x1)>f(x2)14−0.34900.8411−0.34680.8410−0.34540.8411−0.3432−0.8412−0.34680.63f(x1)>f(x2)

Thus, the result converges to true value x=−0.3468.

Answer 15

(b)

Expert Solution

To determine

To calculate: The maximum of the function f(x)=−x4−2x3−8x2−5x with the help of parabolic interpolation. Take initial guesses as x0=−2,x1=−1,x2=1.

Answer to Problem 8P

Solution:

The maximum of the function f(x)=−x4−2x3−8x2−5x is, x=−0.3472 after 4 iterations.

Explanation of Solution

Given information:

The function f(x)=−x4−2x3−8x2−5x. Initial guesses x0=−2,x1=−1,x2=1 and the number of iterations is, 4.

Formula used:

Consider three points jointly bracket an optimum, thus a unique parabola through these three points can be determined. On differentiating and setting it equal to zero estimate of optimal can be computed.

Consider x0,x1 and x2 as initial guesses and x3 be the value of x that corresponds to the maximum value of quadratic which fit into guesses, then,

x3=f(x0)(x12−x22)+f(x1)(x22−x02)+f(x2)(x02−x12)2f(x0)(x1−x2)+2f(x1)(x2−x0)+2f(x2)(x0−x1) …… (1)

Calculation:

Consider function f(x), in standard form as:

f(x)=−x4−2x3−8x2−5x

With initial guesses x0=−2,x1=−1,x2=1.

Iteration 1: Function values at these three initial points is,

For x=−2,

f(x0)=f(−2)=−(−2)4−2(−2)3−8(−2)2−5(−2)=−22

For x=−1,

f(x1)=f(−1)=−(−1)4−2(−1)3−8(−1)2−5(−1)=−2

For x=1,

f(x2)=f(1)=−(1)4−2(1)3−8(1)2−5(1)=−16

Substitute the value of x0,x1,x2 and f(x0),f(x1),f(x2) in equation (1) to get value of x3,

x3=f(x0)(x12−x22)+f(x1)(x22−x02)+f(x2)(x02−x12)2f(x0)(x1−x2)+2f(x1)(x2−x0)+2f(x2)(x0−x1)=(−22)((−1)2−12)+(−2)(12−(−2)2)+(−16)((−2)2−(−1)2)2(−22)((−1)−1)+2(−2)(1−(−2))+2(−16)((−2)−(−1))=−0.3889

The value of function at x3 is,

f(x3)=f(−0.3889)=−(−0.3889)4−2(−0.3889)3−8(−0.3889)2−5(−0.3889)=0.8293

Therefore, f(x3)>f(x1) then the new value of x is to the right of intermediate point x1 and the lower guess x0 is discarded.

Iteration 2: Now the initial guesses are x0=−1,x1=−0.3889,x2=1

Function values at these three initial points are,

For x=−1,

f(x0)=f(−1)=−(−1)4−2(−1)3−8(−1)2−5(−1)=−2

For x=−0.3889,

f(x1)=f(−0.3889)=−(−0.3889)4−2(−0.3889)3−8(−0.3889)2−5(−0.3889)=0.8293

For x=1,

f(x2)=f(1)=−(1)4−2(1)3−8(1)2−5(1)=−16

Substitute the value of x0,x1,x2 and f(x0),f(x1),f(x2) in equation (1) to get value of x3,

x3=f(x0)(x12−x22)+f(x1)(x22−x02)+f(x2)(x02−x12)2f(x0)(x1−x2)+2f(x1)(x2−x0)+2f(x2)(x0−x1)=(−2)((−0.3889)2−12)+(0.8293)(12−(−1)2)+(−16)((−1)2−(−0.3889)2)2(−2)((−0.3889)−1)+2(0.8293)(1−(−1))+2(−16)((−1)−(−0.3889))=−0.4179

The value of the provided function at x3 is,

f(x3)=f(−0.4179)=−(−0.4179)4−2(−0.4179)3−8(−0.4179)2−5(−0.4179)=0.8077

Therefore, f(x1)>f(x3) then the new value of x is to the right of intermediate point x1 and the upper guess x2 is discarded.

Iteration 3: Now the initial guesses are x0=−1,x1=−0.4179,x2=−0.3889

Function values at these three initial points is,

For x=−1,

f(x0)=f(−1)=−(−1)4−2(−1)3−8(−1)2−5(−1)=−2

The function for x1

f(x1)=f(−0.4179)=−(−0.4179)4−2(−0.4179)3−8(−0.4179)2−5(−0.4179)=0.8077

And for x2,

f(x2)=f(−0.3889)=−(−0.3889)4−2(−0.3889)3−8(−0.3889)2−5(−0.3889)=0.8293

Substitute the value of x0,x1,x2 and f(x0),f(x1),f(x2) in equation (1) to get value of x3,

x3=f(x0)(x12−x22)+f(x1)(x22−x02)+f(x2)(x02−x12)2f(x0)(x1−x2)+2f(x1)(x2−x0)+2f(x2)(x0−x1)={(−2)((−0.4179)2−(−0.3889)2)+(0.8077)((−0.3889)2−(−1)2)+(0.8293)((−1)2−(−0.4179)2)}{2(−2)((−0.4179)−(−0.3889))+2(0.8077)((−0.3889)−(−1))+2(0.8293)((−1)−(−0.4179))}=−0.3476

And value of function at x3 is,

f(x3)=f(−0.3476)=−(−0.3476)4−2(−0.3476)3−8(−0.3476)2−5(−0.3476)=0.8408

Therefore, f(x3)>f(x1) then the new value of x is to the right of intermediate point x1 and the lower guess x0 is discarded.

Iteration 4: Now the initial guesses are x0=−0.4179,x1=−0.3889,x2=−0.3476

Function values at these three initial points is,

For x=−0.4179,

f(x0)=f(−0.4179)=−(−0.4179)4−2(−0.4179)3−8(−0.4179)2−5(−0.4179)=0.8077

For x=−0.3889,

f(x1)=f(−0.3889)=−(−0.3889)4−2(−0.3889)3−8(−0.3889)2−5(−0.3889)=0.8293

For x=−0.3476,

f(x2)=f(−0.3476)=−(−0.3476)4−2(−0.3476)3−8(−0.3476)2−5(−0.3476)=0.8408

Substitute the value of x0,x1,x2 and f(x0),f(x1),f(x2) in equation (1) to get value of x3,

x3=(f(x0)(x12−x22)+f(x1)(x22−x02)+f(x2)(x02−x12))(2f(x0)(x1−x2)+2f(x1)(x2−x0)+2f(x2)(x0−x1))=((0.8077)((−0.3889)2−(−0.3476)2)+(0.8293)((−0.3476)2−(−0.4179)2)+(0.8408)((−0.4179)2−(−0.3889)2))(2(0.8077)((−0.3889)−(−0.3476))+2(0.8293)((−0.3476)−(−0.4179))+2(0.8408)((−0.4179)−(−0.3889)))=−0.3472

And value of function at x3 is,

f(x3)=f(−0.3472)=−(−0.3472)4−2(−0.3472)3−8(−0.3472)2−5(−0.3472)=0.8408

Therefore, f(x1)>f(x3) then the new value of x is to the right of intermediate point x1 and the lower guess x0 is discarded.

And the process continues with a summary shown below in a table:

ix0f(x0)x1f(x1)x2f(x2)x3f(x3)εa(%)1−2−22−1−21−16−0.38890.82932−1−2−0.38890.82931−16−0.41790.80776.943−1−2−0.41790.8077−0.38890.8293−0.34760.840820.224−0.41790.8077−0.38890.8293−0.34760.8408−0.34720.84080.115

Thus, after four iterations result is converging to true value x=−0.3472.

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

To determine

To calculate: The maximum of the function f(x)=−x4−2x3−8x2−5x with the help of parabolic interpolation. Take initial guesses as x0=−2,x1=−1,x2=1.

Answer 20

Answer to Problem 8P

Solution:

The maximum of the function f(x)=−x4−2x3−8x2−5x is, x=−0.3472 after 4 iterations.

Answer 21

Explanation of Solution

Given information:

The function f(x)=−x4−2x3−8x2−5x. Initial guesses x0=−2,x1=−1,x2=1 and the number of iterations is, 4.

Formula used:

Consider three points jointly bracket an optimum, thus a unique parabola through these three points can be determined. On differentiating and setting it equal to zero estimate of optimal can be computed.

Consider x0,x1 and x2 as initial guesses and x3 be the value of x that corresponds to the maximum value of quadratic which fit into guesses, then,

x3=f(x0)(x12−x22)+f(x1)(x22−x02)+f(x2)(x02−x12)2f(x0)(x1−x2)+2f(x1)(x2−x0)+2f(x2)(x0−x1) …… (1)

Calculation:

Consider function f(x), in standard form as:

f(x)=−x4−2x3−8x2−5x

With initial guesses x0=−2,x1=−1,x2=1.

Iteration 1: Function values at these three initial points is,

For x=−2,

f(x0)=f(−2)=−(−2)4−2(−2)3−8(−2)2−5(−2)=−22

For x=−1,

f(x1)=f(−1)=−(−1)4−2(−1)3−8(−1)2−5(−1)=−2

For x=1,

f(x2)=f(1)=−(1)4−2(1)3−8(1)2−5(1)=−16

Substitute the value of x0,x1,x2 and f(x0),f(x1),f(x2) in equation (1) to get value of x3,

x3=f(x0)(x12−x22)+f(x1)(x22−x02)+f(x2)(x02−x12)2f(x0)(x1−x2)+2f(x1)(x2−x0)+2f(x2)(x0−x1)=(−22)((−1)2−12)+(−2)(12−(−2)2)+(−16)((−2)2−(−1)2)2(−22)((−1)−1)+2(−2)(1−(−2))+2(−16)((−2)−(−1))=−0.3889

The value of function at x3 is,

f(x3)=f(−0.3889)=−(−0.3889)4−2(−0.3889)3−8(−0.3889)2−5(−0.3889)=0.8293

Therefore, f(x3)>f(x1) then the new value of x is to the right of intermediate point x1 and the lower guess x0 is discarded.

Iteration 2: Now the initial guesses are x0=−1,x1=−0.3889,x2=1

Function values at these three initial points are,

For x=−1,

f(x0)=f(−1)=−(−1)4−2(−1)3−8(−1)2−5(−1)=−2

For x=−0.3889,

f(x1)=f(−0.3889)=−(−0.3889)4−2(−0.3889)3−8(−0.3889)2−5(−0.3889)=0.8293

For x=1,

f(x2)=f(1)=−(1)4−2(1)3−8(1)2−5(1)=−16

Substitute the value of x0,x1,x2 and f(x0),f(x1),f(x2) in equation (1) to get value of x3,

x3=f(x0)(x12−x22)+f(x1)(x22−x02)+f(x2)(x02−x12)2f(x0)(x1−x2)+2f(x1)(x2−x0)+2f(x2)(x0−x1)=(−2)((−0.3889)2−12)+(0.8293)(12−(−1)2)+(−16)((−1)2−(−0.3889)2)2(−2)((−0.3889)−1)+2(0.8293)(1−(−1))+2(−16)((−1)−(−0.3889))=−0.4179

The value of the provided function at x3 is,

f(x3)=f(−0.4179)=−(−0.4179)4−2(−0.4179)3−8(−0.4179)2−5(−0.4179)=0.8077

Therefore, f(x1)>f(x3) then the new value of x is to the right of intermediate point x1 and the upper guess x2 is discarded.

Iteration 3: Now the initial guesses are x0=−1,x1=−0.4179,x2=−0.3889

Function values at these three initial points is,

For x=−1,

f(x0)=f(−1)=−(−1)4−2(−1)3−8(−1)2−5(−1)=−2

The function for x1

f(x1)=f(−0.4179)=−(−0.4179)4−2(−0.4179)3−8(−0.4179)2−5(−0.4179)=0.8077

And for x2,

f(x2)=f(−0.3889)=−(−0.3889)4−2(−0.3889)3−8(−0.3889)2−5(−0.3889)=0.8293

Substitute the value of x0,x1,x2 and f(x0),f(x1),f(x2) in equation (1) to get value of x3,

x3=f(x0)(x12−x22)+f(x1)(x22−x02)+f(x2)(x02−x12)2f(x0)(x1−x2)+2f(x1)(x2−x0)+2f(x2)(x0−x1)={(−2)((−0.4179)2−(−0.3889)2)+(0.8077)((−0.3889)2−(−1)2)+(0.8293)((−1)2−(−0.4179)2)}{2(−2)((−0.4179)−(−0.3889))+2(0.8077)((−0.3889)−(−1))+2(0.8293)((−1)−(−0.4179))}=−0.3476

And value of function at x3 is,

f(x3)=f(−0.3476)=−(−0.3476)4−2(−0.3476)3−8(−0.3476)2−5(−0.3476)=0.8408

Therefore, f(x3)>f(x1) then the new value of x is to the right of intermediate point x1 and the lower guess x0 is discarded.

Iteration 4: Now the initial guesses are x0=−0.4179,x1=−0.3889,x2=−0.3476

Function values at these three initial points is,

For x=−0.4179,

f(x0)=f(−0.4179)=−(−0.4179)4−2(−0.4179)3−8(−0.4179)2−5(−0.4179)=0.8077

For x=−0.3889,

f(x1)=f(−0.3889)=−(−0.3889)4−2(−0.3889)3−8(−0.3889)2−5(−0.3889)=0.8293

For x=−0.3476,

f(x2)=f(−0.3476)=−(−0.3476)4−2(−0.3476)3−8(−0.3476)2−5(−0.3476)=0.8408

Substitute the value of x0,x1,x2 and f(x0),f(x1),f(x2) in equation (1) to get value of x3,

x3=(f(x0)(x12−x22)+f(x1)(x22−x02)+f(x2)(x02−x12))(2f(x0)(x1−x2)+2f(x1)(x2−x0)+2f(x2)(x0−x1))=((0.8077)((−0.3889)2−(−0.3476)2)+(0.8293)((−0.3476)2−(−0.4179)2)+(0.8408)((−0.4179)2−(−0.3889)2))(2(0.8077)((−0.3889)−(−0.3476))+2(0.8293)((−0.3476)−(−0.4179))+2(0.8408)((−0.4179)−(−0.3889)))=−0.3472

And value of function at x3 is,

f(x3)=f(−0.3472)=−(−0.3472)4−2(−0.3472)3−8(−0.3472)2−5(−0.3472)=0.8408

Therefore, f(x1)>f(x3) then the new value of x is to the right of intermediate point x1 and the lower guess x0 is discarded.

And the process continues with a summary shown below in a table:

ix0f(x0)x1f(x1)x2f(x2)x3f(x3)εa(%)1−2−22−1−21−16−0.38890.82932−1−2−0.38890.82931−16−0.41790.80776.943−1−2−0.41790.8077−0.38890.8293−0.34760.840820.224−0.41790.8077−0.38890.8293−0.34760.8408−0.34720.84080.115

Thus, after four iterations result is converging to true value x=−0.3472.

Answer 22

(c)

Expert Solution

To determine

To calculate: The maximum of the function f(x)=−x4−2x3−8x2−5x with the help of Newton’s method. Take initial guess as x0=−1.

Answer to Problem 8P

Solution:

The maximum of the function f(x)=−x4−2x3−8x2−5x is, −0.3472.

Explanation of Solution

Given information:

The function f(x)=−x4−2x3−8x2−5x. Initial guess is x0=−1 and the acceptable error is, et=1%.

Formula used:

Newton Method is similar to Newton Raphson as it does not require initial guesses that bracket the optimum solution.

For any function f(x) optimal solution can be attained by performing iterations following the formula given below:

xi+1=xi−f′(xi)f″(xi) ...... (1)

Calculation:

Consider function f(x),

f(x)=−x4−2x3−8x2−5x

With initial guesses x0=−1,εt=1%.

First and second derivatives of function that is, f′(x) and f″(x), can be evaluated as:

f′(x)= −4x3−6x2−16x−5f″(x)=−12x2−12x−16

Iteration 1:Initially for x0=−1, values of first and second derivatives of function are as follows:

f′(x)= −4x3−6x2−16x−5f′(−1)=−4(−1)3−6(−1)2−16(−1)−5=9

For second derivative,

f″(x)=−12x2−12x−16f″(−1)=−12(−1)2−12(−1)−16=−16

Therefore, x1 is calculated using formula given as equation (2),

x1=x0−f′(x0)f″(x0)=−1−(9)(−16)=−0.4375

And

f(x1)=−x14−2x13−8x12−5x1=−(−0.4375)4−2(−0.4375)3−8(−0.4375)2−5(−0.4375)=0.7871

Iteration 2:Now for x1=−0.4375, values of first and second derivatives of function are as follows:

f′(x1)= −4x13−6x12−16x1−5f′(−0.4375)=−4(−0.4375)3−6(−0.4375)2−16(−0.4375)−5=1.1865

For second derivative,

f″(x)=−12x2−12x−16f″(−0.4375)=−12(−0.4375)2−12(−0.4375)−16=−13.045

Therefore, x2 is calculated using formula given as equation (2)

x2=x1−f′(x1)f″(x1)=−0.4375−(1.1865)(−13.045)=−0.3465

And

f(x2)=−x24−2x23−8x22−5x2=−(−0.3465)4−2(−0.3465)3−8(−0.3465)2−5(−0.3465)=0.8408

Iteration 3:Now for x2=−0.3465, values of first and second derivatives of function are as follows:

f′(x2)= −4x23−6x22−16x2−5f′(−0.3465)=−4(−0.3465)3−6(−0.3465)2−16(−0.3465)−5=−0.0099

For second derivative,

f″(x2)=−12x22−12x2−16f″(−0.3465)=−12(−0.3465)2−12(−0.3465)−16=−13.2827

Therefore, x3 is calculated using formula given as equation (2)

x3=x2−f′(x2)f″(x2)=−0.3465−(−0.0099)(−13.2827)=−0.3472

And

f(x3)=−x34−2x33−8x32−5x3=−(−0.3472)4−2(−0.3472)3−8(−0.3472)2−5(−0.3472)=0.8408

Iteration 4:Now for x3=−0.3472, values of first and second derivatives of function are as follows:

f′(x3)= −4x33−6x32−16x3−5f′(−0.3472)=−4(−0.3472)3−6(−0.3472)2−16(−0.3472)−5=−0.00067

For second derivative,

f″(x2)=−12x22−12x2−16f″(−0.3472)=−12(−0.3472)2−12(−0.3472)−16=−13.2801

Therefore, x4 is calculated using formula given as equation (2)

x4=x3−f′(x3)f″(x3)=−0.3472−(−0.00067)(−13.2801)=−0.347250

Maintaining the error percentage using equation (3) iterations can be summarized as shown in table below:

ixf(x)f′(x)f″(x)εa0−1−29−161−0.43750.78711.865−13.04616.162−0.34650.8408−0.0931−13.2850.113−0.34720.8408−0.0099−13.28000

Thus, within four iterations, the result converges to true value −0.3472.

Therefore, the maximum of the function f(x)=−x4−2x3−8x2−5x is −0.3472.

Answer 23

(c)

Expert Solution

Answer 24

(c)

Expert Solution

Answer 25

Expert Solution

Answer 26

To determine

To calculate: The maximum of the function f(x)=−x4−2x3−8x2−5x with the help of Newton’s method. Take initial guess as x0=−1.

Answer 27

Answer to Problem 8P

Solution:

The maximum of the function f(x)=−x4−2x3−8x2−5x is, −0.3472.

Answer 28

Explanation of Solution

Given information:

The function f(x)=−x4−2x3−8x2−5x. Initial guess is x0=−1 and the acceptable error is, et=1%.

Formula used:

Newton Method is similar to Newton Raphson as it does not require initial guesses that bracket the optimum solution.

For any function f(x) optimal solution can be attained by performing iterations following the formula given below:

xi+1=xi−f′(xi)f″(xi) ...... (1)

Calculation:

Consider function f(x),

f(x)=−x4−2x3−8x2−5x

With initial guesses x0=−1,εt=1%.

First and second derivatives of function that is, f′(x) and f″(x), can be evaluated as:

f′(x)= −4x3−6x2−16x−5f″(x)=−12x2−12x−16

Iteration 1:Initially for x0=−1, values of first and second derivatives of function are as follows:

f′(x)= −4x3−6x2−16x−5f′(−1)=−4(−1)3−6(−1)2−16(−1)−5=9

For second derivative,

f″(x)=−12x2−12x−16f″(−1)=−12(−1)2−12(−1)−16=−16

Therefore, x1 is calculated using formula given as equation (2),

x1=x0−f′(x0)f″(x0)=−1−(9)(−16)=−0.4375

And

f(x1)=−x14−2x13−8x12−5x1=−(−0.4375)4−2(−0.4375)3−8(−0.4375)2−5(−0.4375)=0.7871

Iteration 2:Now for x1=−0.4375, values of first and second derivatives of function are as follows:

f′(x1)= −4x13−6x12−16x1−5f′(−0.4375)=−4(−0.4375)3−6(−0.4375)2−16(−0.4375)−5=1.1865

For second derivative,

f″(x)=−12x2−12x−16f″(−0.4375)=−12(−0.4375)2−12(−0.4375)−16=−13.045

Therefore, x2 is calculated using formula given as equation (2)

x2=x1−f′(x1)f″(x1)=−0.4375−(1.1865)(−13.045)=−0.3465

And

f(x2)=−x24−2x23−8x22−5x2=−(−0.3465)4−2(−0.3465)3−8(−0.3465)2−5(−0.3465)=0.8408

Iteration 3:Now for x2=−0.3465, values of first and second derivatives of function are as follows:

f′(x2)= −4x23−6x22−16x2−5f′(−0.3465)=−4(−0.3465)3−6(−0.3465)2−16(−0.3465)−5=−0.0099

For second derivative,

f″(x2)=−12x22−12x2−16f″(−0.3465)=−12(−0.3465)2−12(−0.3465)−16=−13.2827

Therefore, x3 is calculated using formula given as equation (2)

x3=x2−f′(x2)f″(x2)=−0.3465−(−0.0099)(−13.2827)=−0.3472

And

f(x3)=−x34−2x33−8x32−5x3=−(−0.3472)4−2(−0.3472)3−8(−0.3472)2−5(−0.3472)=0.8408

Iteration 4:Now for x3=−0.3472, values of first and second derivatives of function are as follows:

f′(x3)= −4x33−6x32−16x3−5f′(−0.3472)=−4(−0.3472)3−6(−0.3472)2−16(−0.3472)−5=−0.00067

For second derivative,

f″(x2)=−12x22−12x2−16f″(−0.3472)=−12(−0.3472)2−12(−0.3472)−16=−13.2801

Therefore, x4 is calculated using formula given as equation (2)

x4=x3−f′(x3)f″(x3)=−0.3472−(−0.00067)(−13.2801)=−0.347250

Maintaining the error percentage using equation (3) iterations can be summarized as shown in table below:

ixf(x)f′(x)f″(x)εa0−1−29−161−0.43750.78711.865−13.04616.162−0.34650.8408−0.0931−13.2850.113−0.34720.8408−0.0099−13.28000

Thus, within four iterations, the result converges to true value −0.3472.

Therefore, the maximum of the function f(x)=−x4−2x3−8x2−5x is −0.3472.

Answer 29

Ch. 13 - 13.1 Given the formula (a) Determine the...Ch. 13 - 13.2 Given (a) Plot the function. (b) Use...Ch. 13 - Prob. 3P Ch. 13 - Repeat Prob. 13.3, except use parabolic...Ch. 13 - 13.5 Repeat Prob. 13.3 but use Newton’s method....Ch. 13 - Employ the following methods to find the maximum...Ch. 13 - 13.7 Consider the following function: Use...Ch. 13 - Employ the following methods to find the maximum...Ch. 13 - 13.9 Consider the following function: Perform...Ch. 13 - Consider the following function:...

Concept explainers

Videos

Answer to Problem 8P

Explanation of Solution

Answer to Problem 8P

Explanation of Solution

Answer to Problem 8P

Explanation of Solution

Want to see more full solutions like this?

Chapter 13 Solutions

Additional Math Textbook Solutions