Explanation Given: The given function is, f x , y = x 3 + y 3 − 3 x − 12 y .... 1 Concept used: With the help of D -Test the relative maximum and minimum values of function of two variable can be evaluated with the help of following steps: Evaluate f x , f y , f x x , f y y and f x y and then equate f x and f y equal to 0 to find the critical point P a , b of the function. Evaluate D where D = f x x a , b ⋅ f y y a , b − f x y a , b 2 . At last the theorem states: (a) f has a maximum at point a , b if D > 0 and f x x a , b < 0 . (b) f has a minimum at a , b if D > 0 and f x x a , b > 0 . (c) f has neither a maximum nor a minimum at a , b if D < 0 . The function has a saddle point at a , b . (d) This test is not applicable if D = 0 . Calculation: With the help of concept stated above the maximum and minimum value of the function is evaluated as: Evaluate f x , f y , f x x , f y y and f x y . f x , y = x 3 + y 3 − 3 x − 12 y .... Given f x = 3 x 2 − 3 f y = 3 y 2 − 12 f x x = 6 x f y y = 6 y f x y = 0 Equate f x and f y equal to zero. f x = 0 and f y = 0 3 x 2 − 3 = 0 3 y 2 − 12 = 0 x = − 1 , 1 y = − 2 , 2 So, the ordered pair of critical points are − 1 , − 2 , − 1 , 2 , 1 , − 2 and 1 , 2 for maximum or minimum values. Therefore, evaluating D at these four critical points by using D = f x x a , b ⋅ f y y a , b − f x y a , b 2 . D − 1 , − 2 = f x x − 1 , − 2 ⋅ f y y − 1 , − 2 − f x y − 1 , − 2 2 = − 6 − 12 − 0 = 72 > 0

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ISBN: 9780135164884

Author: BITTINGER

Publisher: PEARSON

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Chapter 6, Problem 9T

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To calculate: The relative maximum and minimum values of fx,y=x3+y3−3x−12y and specify where each occurs.

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The relative maximum and minimum values of f x , y = x 3 + y 3 − 3 x − 12 y and specify where each occurs.

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To calculate: The relative maximum and minimum values of fx,y=x3+y3−3x−12y and specify where each occurs.

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