a. Answer the max value is 144 at x = 2 . Explanation Given information: The function is V ( x ) = x ( 10 − 2 x ) ( 16 − 2 x ) , 0 < x < 5 . Calculation: to find the extreme value, differentiate the function because function has no end points, V ( x ) = x ( 10 − 2 x ) ( 16 − 2 x ) V ( x ) = 12 x 2 − 84 x + 140 [differentiate and simplify] Now, equal the differentiation to zero, 12 x 2 − 84 x + 140 = 0 3 x 2 − 21 x + 35 = 0 [simplify] Substitute the value in the function, V ( x ) = x ( 10 − 2 x ) ( 16 − 2 x ) V ( 2 ) = 2 ( 10 − 4 ) ( 16 − 4 ) [substitute x = 2 ] V ( 2 ) = 144 [simplify] Thus, the max value is 144 at x = 2 . b. To determine To find: the volume by the value occur in part (a). Answer The largest volume of the box is 144 cubic unit and it occur when x = 2 . Explanation Given information: The function is V ( x ) = x ( 10 − 2 x ) ( 16 − 2 x ) , 0 < x < 5 . Calculation: from the part (a), the value is x = 2 an substitute the value of function, V ( x ) = x ( 10 − 2 x ) ( 16 − 2 x ) V ( 2 ) = 2 ( 10 − 4 ) ( 16 − 4 ) [substitute x = 2 ] V ( 2 ) = 144 [simplify] Thus, the max value is 144 at x = 2 . c. To determine To graph: the function. Explanation Given information: The function is V ( x ) = x ( 10 − 2 x ) ( 16 − 2 x ) , 0 < x < 5 . Graph: to graph the function, use the point plotting method, x V ( x ) = x ( 10 − 2 x ) ( 16 − 2 x ) 0 0 2 144 1 112 Now by plotting the points and join the points smoothly, The graph can be obtained as: Interpretation: from the above graph it can be observed that the maximum volume of the box is 144 at x = 2 .

Calculus 2012 Student Edition (by Finney/Demana/Waits/Kennedy)

4th Edition

ISBN: 9780133178579

Author: Ross L. Finney

Publisher: PEARSON

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Question

Chapter 5.1, Problem 43E

To determine

To find: the extreme value of function.

Expert Solution

Answer to Problem 43E

the max value is 144 at x=2 .

Explanation of Solution

Given information:

The function is V(x)=x(10−2x)(16−2x), 0<x<5 .

Calculation: to find the extreme value, differentiate the function because function has no end points,

V(x)=x(10−2x)(16−2x)V(x)=12x2−84x+140 [differentiate and simplify]

Now, equal the differentiation to zero,

12x2−84x+140=03x2−21x+35=0 [simplify]

Substitute the value in the function,

V(x)=x(10−2x)(16−2x)V(2)=2(10−4)(16−4) [substitute x=2]V(2)=144 [simplify]

Thus, the max value is 144 at x=2 .

To determine

To find: the volume by the value occur in part (a).

Expert Solution

Answer to Problem 43E

The largest volume of the box is 144 cubic unit and it occur when x=2 .

Explanation of Solution

Given information:

The function is V(x)=x(10−2x)(16−2x), 0<x<5 .

Calculation: from the part (a), the value is x=2 an substitute the value of function,

V(x)=x(10−2x)(16−2x)V(2)=2(10−4)(16−4) [substitute x=2]V(2)=144 [simplify]

Thus, the max value is 144 at x=2 .

To determine

To graph: the function.

Expert Solution

Explanation of Solution

Given information:

The function is V(x)=x(10−2x)(16−2x), 0<x<5 .

Graph: to graph the function, use the point plotting method,

x	V(x)=x(10−2x)(16−2x)
0	0
2	144
1	112

Now by plotting the points and join the points smoothly,

The graph can be obtained as:

Calculus 2012 Student Edition (by Finney/Demana/Waits/Kennedy), Chapter 5.1, Problem 43E

Interpretation: from the above graph it can be observed that the maximum volume of the box is 144 at x=2 .

Chapter 5.1, Problem 42E

Chapter 5.1, Problem 44E

Chapter 5 Solutions

Calculus 2012 Student Edition (by Finney/Demana/Waits/Kennedy)

Ch. 5.1 - Prob. 1QR Ch. 5.1 - Prob. 2QR Ch. 5.1 - Prob. 3QR Ch. 5.1 - Prob. 4QR Ch. 5.1 - Prob. 5QR Ch. 5.1 - Prob. 6QR Ch. 5.1 - Prob. 7QR Ch. 5.1 - Prob. 8QR Ch. 5.1 - Prob. 9QR Ch. 5.1 - Prob. 10QR

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