Explanation Given: The volume of the rectangular box is 384 cubic inches and inside the box divide by one partition. 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 . Given that, the volume of the box V = 384 . Then, x y z = 384 . …… (1) The total area of the material used to construct the box M ( x , y , z ) is equal to the sum of areas of bottom, sides and partition. That is, M ( x , y , z ) = x y + 2 x z + 3 y z . …… (2) From equation (1), the value of z = 384 x y . Change the function M ( x , y , z ) into two variable by substitute the above z value in the function M ( x , y , z ) . M ( x , y , z ) = x y + 2 x ( 384 x y ) + 3 y ( 384 x y ) = x y + 768 y + 1152 x Calculate the partial derivative of the above function with respect to x . M x ( x , y ) = ∂ ∂ x ( x y + 768 y + 1152 x ) = ∂ ∂ x ( x y ) + ∂ ∂ x ( 768 y ) + ∂ ∂ x ( 1152 x ) = y + 0 − 1152 x 2 = y − 1152 x 2 Again, take the partial derivative of the function M x ( x , y ) with respect to x . M x x ( x , y ) = ∂ ∂ x ( M x ( x , y ) ) = ∂ ∂ x ( y − 1152 x 2 ) = ∂ ∂ x ( y ) − ∂ ∂ x ( 1152 x 2 ) = 2304 x 3 That is, M x x ( x , y ) = 2304 x 3 . …… (3) Calculate the partial derivative of the function M ( x , y ) with respect to y . M y ( x , y ) = ∂ ∂ y ( x y + 768 y + 1152 x ) = ∂ ∂ y ( x y ) + ∂ ∂ y ( 768 y ) + ∂ ∂ y ( 1152 x ) = x − 768 y 2 + 0 = x − 768 y 2 Again, take the partial derivative of the function M y ( x , y ) with respect to y . M y y ( x , y ) = ∂ ∂ y ( M y ( x , y ) ) = ∂ ∂ y ( x − 768 y 2 ) = ∂ ∂ y ( x ) − ∂ ∂ y ( 768 y 2 ) = 1536 y 3 That is, M y y ( x , y ) = 1536 y 3

College Mathematics for Business, Economics, Life Sciences, and Social Sciences (14th Edition)

14th Edition

ISBN: 9780134674148

Author: Raymond A. Barnett, Michael R. Ziegler, Karl E. Byleen, Christopher J. Stocker

Publisher: PEARSON

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Chapter 14.3, Problem 4MP

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

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Chapter 14.3, Problem 3MP

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Chapter 14 Solutions

College Mathematics for Business, Economics, Life Sciences, and Social Sciences (14th Edition)

Ch. 14.1 - Matched Problem 1 Find C(20, 10) for the cost...Ch. 14.1 - Prob. 2MP Ch. 14.1 - Prob. 3MP Ch. 14.1 - Prob. 4MP Ch. 14.1 - Prob. 5MP Ch. 14.1 - Prob. 6MP Ch. 14.1 - Prob. 7MP Ch. 14.1 - Prob. 1ED Ch. 14.1 - Prob. 1E Ch. 14.1 - Prob. 2E

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