Explanation Approach: For a function z = f ( x , y ) , let f x x , f y y , and f x y all exist in a circular region contained in the xy -plane with center ( a , b ) . Further, let f x ( a . b ) = 0 and f y ( a . b ) = 0 . Define the number D , known as the discriminant, by D = f x x ( a , b ) ⋅ f y y ( a , b ) − [ f x y ( a , b ) ] 2 Then, a. f ( a , b ) is a relative maximum if D > 0 and f x x ( a , b ) < 0 . b. f ( a , b ) is a relative minimum if D > 0 and f x x ( a , b ) > 0 . c. f ( a , b ) is a saddle point (neither a maximum nor a minimum) if D < 0 . d. if D = 0 , the test gives no information. Calculation: Consider the given function, f ( x , y ) = 3 2 y − 1 2 y 3 − x 2 y + 1 16 Find f x ( x , y ) , and f y ( x , y ) : f x ( x , y ) = ∂ ∂ x ( 3 2 y − 1 2 y 3 − x 2 y + 1 16 ) = − 2 x y and, f y ( x , y ) = ∂ ∂ y ( 3 2 y − 1 2 y 3 − x 2 y + 1 16 ) = 3 2 − 3 2 y 2 − x 2 A function has relative extrema at a point, if f x ( x , y ) = 0 − 2 x y = 0 And if f y ( x , y ) = 0 , f y ( x , y ) = 0 3 2 − 3 2 y 2 − x 2 = 0 Solve the equation − 2 x y = 0 and 3 2 − 3 2 y 2 − x 2 = 0 using substitution, we get: x = 0 , y = 1 , x = 0 , y = − 1 , x = 3 2 , y = 0 and x = − 3 2 , y = 0 . Therefore, ( 0 , 1 ) , ( 0 , − 1 ) , ( 3 2 , 0 ) and ( − 3 2 , 0 ) are the critical point where extrema can occur. Check for saddle point: f x y ( x , y ) = ∂ ∂ y ( − 2 x y ) = − 2 x , f x x ( x , y ) = ∂ ∂ x ( − 2 x y ) = − 2 y , and, f y y ( x , y ) = ∂ ∂ y ( 3 2 − 3 2 y 2 − x 2 ) = − 3 y Substitute ( 0 , 1 ) in the formula D = f x x ( a , b ) ⋅ f y y ( a , b ) − [ f x y ( a , b ) ] 2 :

Pearson eText for Calculus for the Life Sciences -- Instant Access (Pearson+)

2nd Edition

ISBN: 9780137553457

Author: Raymond Greenwell, Nathan Ritchey

Publisher: PEARSON+

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Chapter 9.3, Problem 22E

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All relative extrema points for the function z=32y−12y3−x2y+116 and then match the equation to its graph for the given function.

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Chapter 9.3, Problem 21E

Chapter 9.3, Problem 23E

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Pearson eText for Calculus for the Life Sciences -- Instant Access (Pearson+)

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All relative extrema points for the function z = 3 2 y − 1 2 y 3 − x 2 y + 1 16 and then match the equation to its graph for the given function.

Solution Summary: The author explains how to find the relative extrema points for the function z=f(x,y) and match the equation to its graph.

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To find:

All relative extrema points for the function z=32y−12y3−x2y+116 and then match the equation to its graph for the given function.

Want to see the full answer?

Chapter 9 Solutions

All relative extrema points for the function z = 3 2 y − 1 2 y 3 − x 2 y + 1 16 and then match the equation to its graph for the given function.

Solution Summary: The author explains how to find the relative extrema points for the function z=f(x,y) and match the equation to its graph.

Concept explainers

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To find: All relative extrema points for the function z=32y−12y3−x2y+116 and then match the equation to its graph for the given function.

Want to see the full answer?

Chapter 9 Solutions

To find:

All relative extrema points for the function z=32y−12y3−x2y+116 and then match the equation to its graph for the given function.