Explanation Result used: Second derivative test: “Suppose C be a critical point at where f ′ ( C ) = 0 , such that f ′ ( x ) exists in a neighbourhood of the critical point C and f ″ ( C ) exists. Then f has a relative maximum value at C if f ″ ( C ) < 0 or relative minimum value at C if f ″ ( C ) > 0 or else the test is not applicable.” Calculation: Obtain the value of x explicitly in terms of y from the function g ( x , y ) = x + 2 y − 9 assuming g ( x , y ) = 0 . x + 2 y − 9 = 0 x = 9 − 2 y Substitute x = 9 − 2 y in the function of f ( x , y ) = 2 x + 3 y − x 2 − y 2 , f ( 9 − 2 y , y ) = 2 ( 9 − 2 y ) + 3 y − ( 9 − 2 y ) 2 − y 2 = 19 − 4 y + 3 y − ( 81 + 4 y 2 − 36 ) − y 2 = 19 − y − 81 − 4 y 2 + 36 − y 2 = − 5 y 2 + 35 y − 63 Consider obtained expression f ( 9 − 2 y , y ) as h ( y ). That is h ( y ) = − 5 y 2 + 35 y − 63 . Compute the derivative of the function h ( y ) = − 5 y 2 + 35 y − 63 . h ′ ( y ) = ( − 5 y 2 + 35 y − 63 ) ′ = − 2 ( 5 y ) + 35 = − 10 y + 35 Substitute h ′ ( y ) = 0 and obtain a value for y , h ′ ( y ) = 0 − 10 y + 35 = 0 − 10 y = − 35 y = − 35 − 10 That is, y = 7 2 is a critical number for the function h

Applied Calculus for the Managerial, Life, and Social Sciences

10th Edition

ISBN: 9781305887831

Author: Tan

Publisher: CENGAGE CO

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Chapter 8.5, Problem 3E

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To maximize: The function f(x,y)=2x+3y−x2−y2 subjects to the constraint of function g(x,y)=x+2y−9 .

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Chapter 8.5, Problem 2E

Chapter 8.5, Problem 4E

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Chapter 8 Solutions

Applied Calculus for the Managerial, Life, and Social Sciences

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To maximize: The function f ( x , y ) = 2 x + 3 y − x 2 − y 2 subjects to the constraint of function g ( x , y ) = x + 2 y − 9 .

Solution Summary: The author explains how to maximize the function f(x,y)=2x+3y-x2- y.

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To maximize: The function f(x,y)=2x+3y−x2−y2 subjects to the constraint of function g(x,y)=x+2y−9 .

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Chapter 8 Solutions