Explanation 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 and the discriminant D = f x x ( a , b ) ⋅ f y y ( a , b ) − [ f x y ( a , b ) ] 2 . (a) The function has relative maximum at ( a , b ) if D > 0 and f x x ( a , b ) < 0 . (b) The function has relative minimum at ( a , b ) if D > 0 and f x x ( a , b ) > 0 . (c) If D < 0 , then f ( a , b ) is a saddle point (neither a maximum nor a minimum) (d) If D = 0 , then the test fail. Calculation: The given time function is, T ( x , y ) = x 4 + 16 y 4 − 32 x y + 40 . Calculate the partial derivative of the above function with respect to x . T x ( x , y ) = ∂ ∂ x ( x 4 + 16 y 4 − 32 x y + 40 ) = ∂ ∂ x ( x 4 ) + ∂ ∂ x ( 16 y 4 ) − ∂ ∂ x ( 32 x y ) + ∂ ∂ x ( 40 ) = 4 x 3 + 0 − 32 y + 0 = 4 x 3 − 32 y Again, take the partial derivative of the function T x ( x , y ) with respect to x . T x x ( x , y ) = ∂ ∂ x ( T x ( x , y ) ) = ∂ ∂ x ( 4 x 3 − 32 y ) = ∂ ∂ x ( 4 x 3 ) − ∂ ∂ x ( 32 y ) = 12 x 2 That is, T x x ( x , y ) = 12 x 2 (1) Calculate the partial derivative of the function T ( x , y ) with respect to y . T y ( x , y ) = ∂ ∂ y ( x 4 + 16 y 4 − 32 x y + 40 ) = ∂ ∂ y ( x 4 ) + ∂ ∂ y ( 16 y 4 ) − ∂ ∂ y ( 32 x y ) + ∂ ∂ y ( 40 ) = 0 + 64 y 3 − 32 x + 0 = 64 y 3 − 32 x Again, take the partial derivative of the function T y ( x , y ) with respect to y . T y y ( x , y ) = ∂ ∂ y ( T y ( x , y ) ) = ∂ ∂ y ( 64 y 3 − 32 x ) = ∂ ∂ y ( 64 y 3 ) − ∂ ∂ y ( 32 x ) = 192 y 2 That is, T y y ( x , y ) = 192 y 2 (2) Calculate the partial derivative of the function T x ( x , y ) with respect to y . T x y ( x , y ) = ∂ ∂ y ( T x ( x , y ) ) = ∂ ∂ y ( 4 x 3 − 32 y ) = ∂ ∂ y ( 4 x 3 ) − ∂ ∂ y ( 32 y ) = − 32 That is, T x y ( x , y ) = − 32 (3) Obtain the critical point by equate the functions T x ( x , y ) = 0 and T y ( x , y ) = 0 . 4 x 3 − 32 y = 0 (4) 64 y 3 − 32 x = 0 (5) From equation (4), find the value of y in terms of x

Finite Mathematics and Calculus with Applications

1st Edition

ISBN: 9781323188361

Author: Margaret Lial

Publisher: Pearson Education

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Chapter 17.3, Problem 39E

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To find: The amount spend on quality control x and amount spent on consulting y such that to minimize the time to spend to meet the quota function T(x,y)=x4+16y4−32xy+40 and use it to find the minimum time of T(x,y) .

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Chapter 17.3, Problem 38E

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The amount spend on quality control x and amount spent on consulting y such that to minimize the time to spend to meet the quota function T ( x , y ) = x 4 + 16 y 4 − 32 x y + 40 and use it to find the minimum time of T ( x , y ) .

Solution Summary: The author explains that the minimum time required to meet the quota set is 8 hours.

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To find: The amount spend on quality control x and amount spent on consulting y such that to minimize the time to spend to meet the quota function T(x,y)=x4+16y4−32xy+40 and use it to find the minimum time of T(x,y) .

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Chapter 17 Solutions