Explanation Given Information: The provided function is: h ( x , y ) = x 2 y − 2 x 2 − 4 y 2 Formula Used: In order to find the critical point of a function, find the partial derivative of the function which respect to each variable and the equate them to zero. Suppose ( a , b ) is a critical point in the interior of the domain of the function f of two variables. Let H be the quantity H = f x x ( a , b ) f y y ( a , b ) − [ f x y ( a , b ) ] 2 , H is called the Hessian. Then, if H is positive, f has a relative minimum at ( a , b ) if f x x ( a , b ) > 0 or f has a relative maximum at ( a , b ) , if f x x ( a , b ) < 0 . If H is negative, f has a saddle point at ( a , b ) , and if H = 0 , the test says nothing. Calculation: Consider the provided function: h ( x , y ) = x 2 y − 2 x 2 − 4 y 2 First, find the first order derivatives h x and h y . To find h x , calculate the partial derivative of the function with respect to x considering y as constant: h x = 2 x y − 4 x …… (1) To find h y , calculate the partial derivative of the function with respect to y considering x as constant: h y = x 2 − 8 y …… (2) Setting the functions h x and h y equal 0 , obtain the critical point as: 2 x y − 4 x = 0 and x 2 − 8 y = 0 Solve the system of equations to obtain the solutions: ( 0 , 0 ) , ( − 4 , 2 ) and ( 4 , 2 ) The critical points are ( 0 , 0 ) , ( − 4 , 2 ) and ( 4 , 2 ) . Now, calculate the values of the second order derivatives h x x , h y y and h x y : Partially differentiate equation (1) with respect to x : h x x = h x ( 2 x y − 4 x ) = 2 y − 4 Partially differentiate equation (2) with respect to y : h y y = h y ( x 2 − 8 y ) = − 8 And partially differentiate equation (1) with respect to y : h x y = h x ( x 2 − 8 y ) = 2 x So, the second order partial derivatives obtained are: h x x = 2 y − 4 , h y y = − 8 and h x y = 2 x …… (3) Now, calculate the Hessian by: H = h x x ( a , b ) h y y ( a , b <

Student Solutions Manual for Waner/Costenoble's Applied Calculus, 7th

7th Edition

ISBN: 9781337291293

Author: Waner, Stefan; Costenoble, Steven

Publisher: Brooks Cole

See similar textbooks

Concept explainers

Domain

In simple mathematics terms, a domain is defined as the space on which all the possible inputs for the function will lie. It can also be expressed as a set of departures for the required function.

Videos

Question

Chapter 8.3, Problem 21E

To determine

To calculate: All the critical points of the function h(x,y)=x2y−2x2−4y2 also locate and classify them.

Expert Solution & Answer

Chapter 8 Solutions

Student Solutions Manual for Waner/Costenoble's Applied Calculus, 7th

Ch. 8.1 - For each function in Exercises 14, evaluate (a)...Ch. 8.1 - For each function in Exercises 14, evaluate (a)...Ch. 8.1 - For each function in Exercises 14, evaluate (a)...Ch. 8.1 - For each function in Exercises 14, evaluate (a)...Ch. 8.1 - For each function in Exercises 5-8, evaluate (a)...Ch. 8.1 - Prob. 6E Ch. 8.1 - For each function in Exercises 5-8, evaluate (a)...Ch. 8.1 - For each function in Exercises 5-8, evaluate (a)...Ch. 8.1 - Prob. 9E Ch. 8.1 - Prob. 10E

Knowledge Booster

$Calculus$

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, calculus and related others by exploring similar questions and additional content below.

All the critical points of the function h ( x , y ) = x 2 y − 2 x 2 − 4 y 2 also locate and classify them.

Solution Summary: The author explains how to calculate the critical points of the function h(x,y) and the partial derivative of it.

Concept explainers

Videos

To calculate: All the critical points of the function h(x,y)=x2y−2x2−4y2 also locate and classify them.

Chapter 8 Solutions