Explanation Result used: Second Derivative Test: “Suppose the second partial derivatives of f are continuous on a disk with center ( a , b ) , and suppose that f x ( a , b ) = 0 and f y ( a , b ) = 0 (that is ( a , b ) is a critical point of f ). Let D ( a , b ) = f x x ( a , b ) f y y ( a , b ) − [ f x y ( a , b ) ] 2 ( a ) If D ( a , b ) > 0 and f x x ( a , b ) > 0 , then f ( a , b ) is a local minimum. ( b ) If D ( a , b ) > 0 and f x x ( a , b ) < 0 , then f ( a , b ) is a local maximum. ( c ) If D ( a , b ) < 0 , then f ( a , b ) is a saddle point. ( d ) If D ( a , b ) = 0 , then the test is inconclusive”. Calculation: The given function is, f ( x , y ) = 4 + 2 x 2 + 3 y 2 . Take the partial derivative in the given function with respect to x and obtain f x . ∂ f ∂ x = ∂ ∂ x ( 4 + 2 x 2 + 3 y 2 ) = ∂ ∂ x ( 4 ) + ∂ ∂ x ( 2 x 2 ) + ∂ ∂ x ( 3 y 2 ) = 0 + 2 ( 2 x ) + 0 = 4 x Thus, ∂ f ∂ x = 4 x (1) Take the partial derivative in the given function with respect to y and obtain f y . ∂ f ∂ y = ∂ ∂ y ( 4 + 2 x 2 + 3 y 2 ) = ∂ ∂ y ( 4 ) + ∂ ∂ y ( 2 x 2 ) + ∂ ∂ y ( 3 y 2 ) = 0 + 0 + 3 ( 2 y ) = 6 y Thus, ∂ f ∂ y = 6 y (2) Set the above partial derivatives to 0 and find the values of x and y . From the equation (1), 4 x = 0 x = 0 From the equation (2), 6 y = 0 y = 0 Thus, the critical point is, ( 0 , 0 )

Calculus: Early Transcendentals (3rd Edition)

3rd Edition

ISBN: 9780134763644

Author: William L. Briggs, Lyle Cochran, Bernard Gillett, Eric Schulz

Publisher: PEARSON

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Question

Chapter 15.7, Problem 25E

To determine

To find: The critical points for the function f(x,y)=4+2x2+3y2 and use second derivative test to check the local maximum, local minimum and saddle point.

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Chapter 15.7, Problem 24E

Chapter 15.7, Problem 26E

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