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 function is, z = 3 2 y − 1 2 y 3 − x 2 y + 1 16 . Calculate the partial derivative of the above function with respect to x . z x = ∂ ∂ x ( 3 2 y − 1 2 y 3 − x 2 y + 1 16 ) = ∂ ∂ x ( 3 2 y ) − ∂ ∂ x ( 1 2 y 3 ) − ∂ ∂ x ( x 2 y ) + ∂ ∂ x ( 1 16 ) = 0 + 0 − 2 x y + 0 = − 2 x y Again, take the partial derivative of the function z x with respect to x . z x x = ∂ ∂ x ( z x ) = ∂ ∂ x ( − 2 x y ) = − 2 ∂ ∂ x ( x y ) = − 2 y That is, z x x = − 2 y (1) Calculate the partial derivative of the function z with respect to y . z y = ∂ ∂ y ( 3 2 y − 1 2 y 3 − x 2 y + 1 16 ) = ∂ ∂ y ( 3 2 y ) − ∂ ∂ y ( 1 2 y 3 ) − ∂ ∂ y ( x 2 y ) + ∂ ∂ y ( 1 16 ) = 3 2 − 3 2 y 2 − x 2 + 0 = 3 2 − 3 2 y 2 − x 2 Again, take the partial derivative of the function z y with respect to y . z y y = ∂ ∂ y ( z y ) = ∂ ∂ y ( 3 2 − 3 2 y 2 − x 2 ) = ∂ ∂ y ( 3 2 ) − ∂ ∂ y ( 3 2 y 2 ) − ∂ ∂ y ( x 2 ) = − 3 y That is, z y y = − 3 y (2) Calculate the partial derivative of the function z x with respect to y . z x y = ∂ ∂ y ( z x ) = ∂ ∂ y ( − 2 x y ) = − 2 x That is, z x y = − 2 x (3) Obtain the critical point by equate the functions z x = 0 and z y = 0 . − 2 x y = 0 (4) 3 2 − 3 2 y 2 − x 2 = 0 (5) From equations (4), find the value of y and x . − 2 x y = 0 x y = 0 x = 0 or y = 0 Substitute the value x = 0 in the equation (5) and find the respective values of y . 3 2 − 3 2 y 2 − ( 0 ) 2 = 0 3 2 ( 1 − y 2 ) = 0 1 − y 2 = 0 y 2 = 1 y = 1 and y = − 1 Therefore, the corresponding value of y for x = 0 is 1 and −1. Substitute the value y = 0 in the equation (5) and find the respective values of x . 3 2 − 3 2 ( 0 ) 2 − x 2 = 0 3 2 − x 2 = 0 x 2 = 3 2 x = 3 2 and x = − 3 2 Therefore, the corresponding value of x for y = 0 is 6 2 and − 6 2 . Thus, the critical point of the function are ( 0 , 1 ) , ( 0 , − 1 ) , ( 6 2 , 0 ) and ( − 6 2 , 0 ) . The discriminant D at the point ( 0 , 1 ) is calculated as follows. By (1), obtain the value z x x ( 0 , 1 ) . f x x ( 0 , 1 ) = − 2 ( 1 ) = − 2 Therefore, the value z x x ( 0 , 1 ) = − 2 . By (2), obtain the value z y y ( 0 , 1 ) . z y y ( 0 , 1 ) = − 3 ( 1 ) = − 3 Thus, the value of z y y ( 0 , 1 ) = − 3 . By (3), obtain the value z x y ( 0 , 1 ) . z x y ( 0 , 1 ) = − 2 ( 0 ) = 0 Thus, the value of z x y ( 0 , 1 ) = 0 . Now, the value of the discriminant is, D = z x x ( 0 , 1 ) ⋅ z y y ( 0 , 1 ) − [ z x y ( 0 , 1 ) ] 2 = ( − 2 ) ⋅ ( − 3 ) − ( 0 ) 2 = 6 − 0 = 6 Thus, the discriminant value D > 0 at the point ( 0 , 1 ) and also the value z x x ( 0 , 1 ) < 0 . By result (a) , the function has a relative maximum at the point ( 0 , 1 ) . Obtain the relative maximum value of the function z = 3 2 y − 1 2 y 3 − x 2 y + 1 16 as follows. Substitute the value x = 0 and y = 1 in the equation z = 3 2 y − 1 2 y 3 − x 2 y + 1 16

Finite Mathematics and Calculus with Applications Books a la Carte Plus MyLab Math Package (10th Edition)

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ISBN: 9780133935592

Author: Margaret L. Lial, Raymond N. Greenwell, Nathan P. Ritchey

Publisher: PEARSON

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Chapter 17.3, Problem 22E

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To find: The set of all relative extrema of the function z=32y−12y3−x2y+116 and use the relative extrema to match the function form the given graph (a) to (f).

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Chapter 17.3, Problem 21E

Chapter 17.3, Problem 23E

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Finite Mathematics and Calculus with Applications Books a la Carte Plus MyLab Math Package (10th Edition)

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If the function...Ch. 17 - Prob. 85RE Ch. 17 - Prob. 86RE Ch. 17 - Prob. 87RE Ch. 17 - Prob. 88RE Ch. 17 - Prob. 89RE Ch. 17 - Prob. 90RE Ch. 17 - Prob. 91RE Ch. 17 - Prob. 92RE Ch. 17 - Prob. 93RE Ch. 17 - Prob. 94RE Ch. 17 - Prob. 95RE Ch. 17 - Prob. 96RE Ch. 17 - Profit The total profit from 1 acre of a certain...Ch. 17 - Prob. 98RE Ch. 17 - Prob. 99RE Ch. 17 - Prob. 100RE Ch. 17 - Prob. 101RE Ch. 17 - Prob. 102RE Ch. 17 - Prob. 103RE Ch. 17 - Prob. 104RE Ch. 17 - Prob. 105RE Ch. 17 - Prob. 106RE

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The set of all relative extrema of the function z = 3 2 y − 1 2 y 3 − x 2 y + 1 16 and use the relative extrema to match the function form the given graph (a) to (f).

Solution Summary: The author explains the relative extrema of the function z=32y - f_xx, and the relation between the functions.

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To find: The set of all relative extrema of the function z=32y−12y3−x2y+116 and use the relative extrema to match the function form the given graph (a) to (f).

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