a. Answer A = 8 x − ( π + 4 8 ) x 2 Explanation Calculation: A window of perimeter 16 feet is constructed by adjoining a semicircle to the top of an ordinary rectangular window is given below: The perimeter of the window is: Perimeter = Length of three sides of the rectangle + arc of the semi circle P = x + y + y + 1 2 ( 2 π r ) = 2 y + x + π ( x 2 ) = 2 y + ( 1 + π 2 ) x It is given that the perimeter of window is16. This gives that, P = 16 2 y + ( 1 + π 2 ) x = 16 2 y = 16 − ( 1 + π 2 ) x y = 16 − ( 2 + π 2 ) x 2 y = 8 − ( 2 + π 4 ) x Now area of the window is, Area = Area of rectangle + Area of semi circle A = x y + 1 2 π ( x 2 ) 2 = x y + π 8 x 2 Substitute y = 8 − ( 2 + π 4 ) x in the above area. A = x y + π 2 x 2 = x ( 8 − ( 2 + π 4 ) x ) + π 8 x 2 = 8 x + ( π 8 − 2 + π 4 ) x 2 = 8 x − ( π + 4 8 ) x 2 Conclusion The area A of the window as a function of x is A = 8 x − ( π + 4 8 ) x 2 . b. To determine To find: the dimensions that produce maximum enclosed area. Answer The dimensions that produce maximum enclosed area are x = 4.48 ft and y = 2.24 ft . Explanation Given: The area A of the window as a function of x is A = 8 x − ( π + 4 8 ) x 2 Concept Used: For the function f ( x ) = a x 2 + b x + c with vertex ( − b 2 a , f ( − b 2 a ) ) , When a > 0 , f has a minimum at x = − b 2 a and the minimum value is f ( − b 2 a ) . When a < 0 , f has a maximum at x = − b 2 a and the maximum value is f ( − b 2 a ) . Calculation: Now compare the given function A = 8 x − ( π + 4 8 ) x 2 with f ( x ) = a x 2 + b x + c . Clearly a = − ( π + 4 8 ) < 0 and b = 8 So by the above definition, f has a maximum at x = − b 2 a and the maximum value is f ( − b 2 a ) . So, the dimensions that produce maximum enclosed area is, x = − b 2 a = − 8 2 × ( − ( π + 4 8 ) ) = 32 π + 4 ≈ 4.48 From part (a), y = 8 − ( 2 + π 4 ) ( 4.48 ) ≈ 2.24 Conclusion: The dimensions that produce maximum enclosed area are x = 4.48 ft and y = 2.24 ft .

4th Edition

ISBN: 9781337516853

Author: Larson

Publisher: CENGAGE CO

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Question

Chapter 2.1, Problem 74E

To determine

To find: the area A of the window as a function of x .

Expert Solution

Answer to Problem 74E

A=8x−(π+48)x2

Explanation of Solution

Calculation:

A window of perimeter 16 feet is constructed by adjoining a semicircle to the top of an ordinary rectangular window is given below:

EBK PRECALCULUS W/LIMITS, Chapter 2.1, Problem 74E

The perimeter of the window is:

Perimeter = Length of three sides of the rectangle + arc of the semi circle

P=x+y+y+12(2πr)=2y+x+π(x2)=2y+(1+π2)x

It is given that the perimeter of window is16. This gives that,

P=162y+(1+π2)x=162y=16−(1+π2)xy=16−(2+π2)x2

y=8−(2+π4)x

Now area of the window is,

Area = Area of rectangle + Area of semi circle

A=xy+12π(x2)2=xy+π8x2

Substitute y=8−(2+π4)x in the above area.

A=xy+π2x2=x(8−(2+π4)x)+π8x2=8x+(π8−2+π4)x2=8x−(π+48)x2

Conclusion The area A of the window as a function of x is A=8x−(π+48)x2 .

To determine

To find: the dimensions that produce maximum enclosed area.

Expert Solution

Answer to Problem 74E

The dimensions that produce maximum enclosed area are x=4.48 ft and y=2.24 ft .

Explanation of Solution

Given:

The area A of the window as a function of x is

A=8x−(π+48)x2

Concept Used:

For the function f(x)=ax2+bx+c with vertex (−b2a,f(−b2a)) ,

When a>0 , f has a minimum at x=−b2a and the minimum value is f(−b2a) .

When a<0 , f has a maximum at x=−b2a and the maximum value is f(−b2a) .

Calculation:

Now compare the given function A=8x−(π+48)x2 with f(x)=ax2+bx+c .

Clearly a=−(π+48)<0 and b=8

So by the above definition, f has a maximum at x=−b2a and the maximum value is f(−b2a) .

So, the dimensions that produce maximum enclosed area is,

x=−b2a=−82×(−(π+48))=32π+4≈4.48

From part (a),

y=8−(2+π4)(4.48)≈2.24

Conclusion:

The dimensions that produce maximum enclosed area are x=4.48 ft and y=2.24 ft .

Chapter 2.1, Problem 73E

Chapter 2.1, Problem 75E

Chapter 2 Solutions

EBK PRECALCULUS W/LIMITS

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Prob. 64E Ch. 2.6 - Prob. 65E Ch. 2.6 - Prob. 66E Ch. 2.6 - Prob. 67E Ch. 2.6 - Prob. 68E Ch. 2.6 - Prob. 69E Ch. 2.6 - Prob. 70E Ch. 2.6 - Prob. 71E Ch. 2.6 - Prob. 72E Ch. 2.6 - Prob. 73E Ch. 2.6 - Prob. 74E Ch. 2.6 - Prob. 75E Ch. 2.6 - Prob. 76E Ch. 2.6 - Prob. 77E Ch. 2.6 - Prob. 78E Ch. 2.6 - Prob. 79E Ch. 2.6 - Prob. 80E Ch. 2.6 - Prob. 81E Ch. 2.6 - Prob. 82E Ch. 2.7 - Prob. 1E Ch. 2.7 - Prob. 2E Ch. 2.7 - Prob. 3E Ch. 2.7 - Prob. 4E Ch. 2.7 - Prob. 5E Ch. 2.7 - Prob. 6E Ch. 2.7 - Prob. 7E Ch. 2.7 - Prob. 8E Ch. 2.7 - Prob. 9E Ch. 2.7 - Prob. 10E Ch. 2.7 - Prob. 11E Ch. 2.7 - Prob. 12E Ch. 2.7 - Prob. 13E Ch. 2.7 - Prob. 14E Ch. 2.7 - Prob. 15E Ch. 2.7 - Prob. 16E Ch. 2.7 - Prob. 17E Ch. 2.7 - Prob. 18E Ch. 2.7 - Prob. 19E Ch. 2.7 - Prob. 20E Ch. 2.7 - Prob. 21E Ch. 2.7 - Prob. 22E Ch. 2.7 - Prob. 23E Ch. 2.7 - Prob. 24E Ch. 2.7 - Prob. 25E Ch. 2.7 - Prob. 26E Ch. 2.7 - Prob. 27E Ch. 2.7 - Prob. 28E Ch. 2.7 - Prob. 29E Ch. 2.7 - Prob. 30E Ch. 2.7 - Prob. 31E Ch. 2.7 - Prob. 32E Ch. 2.7 - Prob. 33E Ch. 2.7 - Prob. 34E Ch. 2.7 - Prob. 35E Ch. 2.7 - Prob. 36E Ch. 2.7 - Prob. 37E Ch. 2.7 - Prob. 38E Ch. 2.7 - Prob. 39E Ch. 2.7 - Prob. 40E Ch. 2.7 - Prob. 41E Ch. 2.7 - Prob. 42E Ch. 2.7 - Prob. 43E Ch. 2.7 - Prob. 44E Ch. 2.7 - Prob. 45E Ch. 2.7 - Prob. 46E Ch. 2.7 - Prob. 47E Ch. 2.7 - Prob. 48E Ch. 2.7 - Prob. 49E Ch. 2.7 - Prob. 50E Ch. 2.7 - Prob. 51E Ch. 2.7 - Prob. 52E Ch. 2.7 - Prob. 53E Ch. 2.7 - Prob. 54E Ch. 2.7 - Prob. 55E Ch. 2.7 - Prob. 56E Ch. 2.7 - Prob. 57E Ch. 2.7 - Prob. 58E Ch. 2.7 - Prob. 59E Ch. 2.7 - Prob. 60E Ch. 2.7 - Prob. 61E Ch. 2.7 - Prob. 62E Ch. 2.7 - Prob. 63E Ch. 2.7 - Prob. 64E Ch. 2.7 - Prob. 65E Ch. 2.7 - Prob. 66E Ch. 2.7 - Prob. 67E Ch. 2.7 - Prob. 68E Ch. 2.7 - Prob. 69E Ch. 2.7 - Prob. 70E Ch. 2.7 - Prob. 71E Ch. 2.7 - Prob. 72E Ch. 2.7 - Prob. 73E Ch. 2.7 - Prob. 74E Ch. 2.7 - Prob. 75E Ch. 2.7 - Prob. 76E Ch. 2.7 - Prob. 77E Ch. 2.7 - Prob. 78E Ch. 2.7 - Prob. 79E Ch. 2.7 - Prob. 80E Ch. 2.7 - Prob. 81E Ch. 2.7 - Prob. 82E Ch. 2.7 - Prob. 83E Ch. 2.7 - Prob. 84E Ch. 2.7 - Prob. 85E Ch. 2.7 - Prob. 86E Ch. 2.7 - Prob. 87E Ch. 2.7 - Prob. 88E Ch. 2.7 - Prob. 89E Ch. 2.7 - Prob. 90E Ch. 2 - Prob. 1RE Ch. 2 - Prob. 2RE Ch. 2 - Prob. 3RE Ch. 2 - Prob. 4RE Ch. 2 - Prob. 5RE Ch. 2 - Prob. 6RE Ch. 2 - Prob. 7RE Ch. 2 - Prob. 8RE Ch. 2 - Prob. 9RE Ch. 2 - Prob. 10RE Ch. 2 - Prob. 11RE Ch. 2 - Prob. 12RE Ch. 2 - Prob. 13RE Ch. 2 - Prob. 14RE Ch. 2 - Prob. 15RE Ch. 2 - Prob. 16RE Ch. 2 - Prob. 17RE Ch. 2 - Prob. 18RE Ch. 2 - Prob. 19RE Ch. 2 - Prob. 20RE Ch. 2 - Prob. 21RE Ch. 2 - Prob. 22RE Ch. 2 - Prob. 23RE Ch. 2 - Prob. 24RE Ch. 2 - Prob. 25RE Ch. 2 - Prob. 26RE Ch. 2 - Prob. 27RE Ch. 2 - Prob. 28RE Ch. 2 - Prob. 29RE Ch. 2 - Prob. 30RE Ch. 2 - Prob. 31RE Ch. 2 - Prob. 32RE Ch. 2 - Prob. 33RE Ch. 2 - Prob. 34RE Ch. 2 - Prob. 35RE Ch. 2 - Prob. 36RE Ch. 2 - Prob. 37RE Ch. 2 - Prob. 38RE Ch. 2 - Prob. 39RE Ch. 2 - Prob. 40RE Ch. 2 - Prob. 41RE Ch. 2 - Prob. 42RE Ch. 2 - Prob. 43RE Ch. 2 - Prob. 44RE Ch. 2 - Prob. 45RE Ch. 2 - Prob. 46RE Ch. 2 - Prob. 47RE Ch. 2 - Prob. 48RE Ch. 2 - Prob. 49RE Ch. 2 - Prob. 50RE Ch. 2 - Prob. 51RE Ch. 2 - Prob. 52RE Ch. 2 - Prob. 53RE Ch. 2 - Prob. 54RE Ch. 2 - Prob. 55RE Ch. 2 - Prob. 56RE Ch. 2 - Prob. 57RE Ch. 2 - Prob. 58RE Ch. 2 - Prob. 59RE Ch. 2 - Prob. 60RE Ch. 2 - Prob. 61RE Ch. 2 - Prob. 62RE Ch. 2 - Prob. 63RE Ch. 2 - Prob. 64RE Ch. 2 - Prob. 65RE Ch. 2 - Prob. 66RE Ch. 2 - Prob. 67RE Ch. 2 - Prob. 68RE Ch. 2 - Prob. 1CT Ch. 2 - Prob. 2CT Ch. 2 - Prob. 3CT Ch. 2 - Prob. 4CT Ch. 2 - Prob. 5CT Ch. 2 - Prob. 6CT Ch. 2 - Prob. 7CT Ch. 2 - Prob. 8CT Ch. 2 - Prob. 9CT Ch. 2 - Prob. 10CT Ch. 2 - Prob. 11CT Ch. 2 - Prob. 12CT Ch. 2 - Prob. 13CT Ch. 2 - Prob. 14CT Ch. 2 - Prob. 15CT Ch. 2 - Prob. 16CT Ch. 2 - Prob. 17CT Ch. 2 - Prob. 18CT Ch. 2 - Prob. 1PS Ch. 2 - Prob. 2PS Ch. 2 - Prob. 3PS Ch. 2 - Prob. 4PS Ch. 2 - Prob. 5PS Ch. 2 - Prob. 6PS Ch. 2 - Prob. 7PS Ch. 2 - Prob. 8PS Ch. 2 - Prob. 9PS Ch. 2 - Prob. 10PS Ch. 2 - Prob. 11PS Ch. 2 - Prob. 12PS Ch. 2 - Prob. 13PS

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