Explanation Given: The volume of the rectangular box is 96 cubic inches. The box divide by two vertical partitions and horizontal partitions. Result used: “Let the partial derivatives f x x , f y y and f x y of the function z = f ( x , y ) exists in the circular region contained in the xy -plane with center ( a , b ) and f x ( a , b ) = 0 , f y ( a , b ) = 0 , then (a) The function has relative maximum at ( a , b ) if f x x ( a , b ) ⋅ f y y ( a , b ) − [ f x y ( a , b ) ] 2 > 0 and f x x ( a , b ) < 0 . (b) The function has relative minimum at ( a , b ) if f x x ( a , b ) ⋅ f y y ( a , b ) − [ f x y ( a , b ) ] 2 > 0 and f x x ( a , b ) > 0 . (c) If f x x ( a , b ) ⋅ f y y ( a , b ) − [ f x y ( a , b ) ] 2 < 0 , then f ( a , b ) is a saddle point (neither a maximum nor a minimum) (d) If f x x ( a , b ) ⋅ f y y ( a , b ) − [ f x y ( a , b ) ] 2 = 0 , then the test fails.” Calculation: Let x , y and z be the length, width and height of the rectangular box, respectively. Then, the volume of the rectangular box is V = x y z . Since the volume of the box V = 96 , x y z = 96 (1) The total surface area S ( x , y , z ) of the rectangular box is equal to area of the bottom plus sides plus vertical partitions plus horizontal partition. That is, S ( x , y , z ) = x y + 2 x z + 2 y z + 2 x z + y z (2) From equation (1), the value of z = 96 x y . Change the function S ( x , y , z ) into two variable by substitute the above z value in the function S ( x , y , z ) . S ( x , y ) = x y + 4 x ( 96 x y ) + 3 y ( 96 x y ) = x y + 384 y + 288 x Calculate the partial derivative of the above function with respect to x . S x ( x , y ) = ∂ ∂ x ( x y + 384 y + 288 x ) = ∂ ∂ x ( x y ) + ∂ ∂ x ( 384 y ) + ∂ ∂ x ( 288 x ) = y + 0 − 288 x 2 = y − 288 x 2 Again, take the partial derivative of the function S x ( x , y ) with respect to x . S x x ( x , y ) = ∂ ∂ x ( S x ( x , y ) ) = ∂ ∂ x ( y − 288 x 2 ) = ∂ ∂ x ( y ) − ∂ ∂ x ( 288 x 2 ) = 576 x 3 That is, S x x ( x , y ) = 576 x 3 (3) Calculate the partial derivative of the function S ( x , y ) with respect to y . S y ( x , y ) = ∂ ∂ y ( x y + 384 y + 288 x ) = ∂ ∂ y ( x y ) + ∂ ∂ y ( 384 y ) + ∂ ∂ y ( 288 x ) = x − 384 y 2 + 0 = x − 384 y 2 Again, take the partial derivative of the function S y ( x , y ) with respect to y . S y y ( x , y ) = ∂ ∂ y ( S y ( x , y ) ) = ∂ ∂ y ( x − 384 y 2 ) = ∂ ∂ y ( x ) − ∂ ∂ y ( 384 y 2 ) = 768 y 3 That is, S y y ( x , y ) = 768 y 3 (4) Calculate the partial derivative of the function S x ( x , y ) with respect to y

Calculus for Business, Economics, Life Sciences, and Social Sciences - Boston U.

15th Edition

ISBN: 9781323047620

Author: Barnett, Ziegler, Byleen

Publisher: Pearson Education

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Chapter 7, Problem 26RE

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To find: The dimensions of the rectangular box that will require the least material.

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Chapter 7, Problem 25RE

Chapter 7, Problem 27RE

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Calculus for Business, Economics, Life Sciences, and Social Sciences - Boston U.

Ch. 7.1 - Prob. 1MP Ch. 7.1 - Matched Problem 2 Find f(2, 2, 3) for f in Example...Ch. 7.1 - Prob. 3MP Ch. 7.1 - Matched Problem 4 For the box in Example 4, find...Ch. 7.1 - Matched Problem 5 Refer to Example 5. Find the...Ch. 7.1 - Matched Problem 6 Find the coordinates of the...Ch. 7.1 - Matched Problem 7 (A) Describe the cross sections...Ch. 7.1 - Explore and Discuss 1 Imagine that you are facing...Ch. 7.1 - In Problems 18, find the indicated value of the...Ch. 7.1 - Prob. 2E

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The dimensions of the rectangular box that will require the least material.

Solution Summary: The author explains the dimensions of the rectangular box that will require the least material. The box divides by two vertical and horizontal partitions.

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Chapter 7 Solutions