Explanation Given: The function is, f ( x , y ) = − x 2 − y 2 + 10 x + 12 y − 64 . Formula used: The extrema will occur based on these conditions: If d = z x x z y y − ( z x y ) 2 > 0 , and z x x > 0 , then minima will occur at its critical point. If d = z x x z y y − ( z x y ) 2 > 0 , and z x x < 0 , then maxima will occur at its critical point. Calculation: Consider the expression, f ( x , y ) = − x 2 − y 2 + 10 x + 12 y − 64 . Now, rewrite f ( x , y ) in the perfect square form, f ( x , y ) = − x 2 + 10 x − 25 − y 2 + 12 y − 36 − 3 = − ( x 2 − 10 x + 25 ) − ( y 2 − 12 y + 36 ) − 3 = − ( x − 5 ) 2 − ( y − 6 ) 2 − 3 A square function has a minimum value of 0. Then, ( x − 5 ) 2 is minimum at x = 5 . Similarly, ( y − 6 ) 2 is minimum at y = 6 >. Hence, for function f ( x , y ) = − x 2 − y 2 + 10 x + 12 y − 64 maximum occurs at ( 5 , 6 , − 3 ) . To verify this result, partial differentiate f ( x , y ) = − x 2 − y 2 + 10 x + 12 y − 64 with respect to x and y

Calculus: Early Transcendental Functions (MindTap Course List)

6th Edition

ISBN: 9781285774770

Author: Ron Larson, Bruce H. Edwards

Publisher: Cengage Learning

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Chapter 13.8, Problem 6E

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To calculate: The extrema of the function f(x,y)=−x2−y2+10x+12y−64 by recognizing the given form. Verify this result by partial derivatives to locate any critical points and test for relative extrema.

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Chapter 13.8, Problem 5E

Chapter 13.8, Problem 7E

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The extrema of the function f ( x , y ) = − x 2 − y 2 + 10 x + 12 y − 64 by recognizing the given form. Verify this result by partial derivatives to locate any critical points and test for relative extrema.

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To calculate: The extrema of the function f(x,y)=−x2−y2+10x+12y−64 by recognizing the given form. Verify this result by partial derivatives to locate any critical points and test for relative extrema.

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Chapter 13 Solutions