Download the App!

Get 24/7 study help with the Numerade app for iOS and Android! Enter your email for an invite.

Get the answer to your homework problem.

Try Numerade free for 7 days

Like

Report

Find and classify, using the second partial derivative test, the critical points of the function defined .$$f(x, y)=-4 x^{2} y-3 x y^{2}+48 x y$$

$$\operatorname{sp}(0,0,0), \operatorname{sp}((12,0,0), \operatorname{sp}(0,16,0), \operatorname{rM}(4,16 / 3,1024 / 3)$$

Calculus 3

Chapter 6

An Introduction to Functions of Several Variables

Section 3

Extrema

Partial Derivatives

Johns Hopkins University

Oregon State University

Baylor University

University of Nottingham

Lectures

12:15

In calculus, partial derivatives are derivatives of a function with respect to one or more of its arguments, where the other arguments are treated as constants. Partial derivatives contrast with total derivatives, which are derivatives of the total function with respect to all of its arguments.

04:14

02:01

Find and classify, using t…

01:48

01:22

02:03

02:48

01:29

02:18

03:06

01:23

01:39

01:51

01:57

02:34

01:55

01:59

for this problem we are asked to find and classify using the second partial derivative. Test the critical points of the function F of x. Y equals negative four X squared y minus three X Y squared plus 48 X. Y. To begin, we want to find the critical points by finding where the gradient of our function will equal zero. So the gradient of our function is going to be 48 y minus eight xy minus three Y squared in the first component and 48 x minus four X square -6 X. Y. In the second component. If we want that to equal our zero vector then we have a few different possible solutions are first solution Is the .016. The second, The .416/3. Our third The point of 12 0 And our 4th is the zero. So having our critical points, we can now proceed with applying our second derivative test but we need to actually take the second partial derivatives first. So the second partial with respect to X is going to be negative eight Y. Second partial why Is going to be -6 X. And the mixed partial is going to be 48 -8, X -6 Y. Having that then our d function that we use for testing the f x Y. Is going to be 48 xy minus minus 48 -8 X -6 Y. All squared. Now Testing the .016 We'll get a result of negative 2304 Which is less than zero, indicating that 016 actual noted up above the saddle point testing. Then 4, 16/3 We have a result of 768 is greater than zero. And we can see that the second partial derivative would be negative at that point. Our second partial derivative with respect to X will be negative at that point, which means that we'll have a relative max at the point for 16/3. At the .12 0 We have negative 2304 Less than zero indicating another saddle. And at the zero We get another value of negative 2304 Less than zero, which means that the zero is a saddle point as well.

View More Answers From This Book

Find Another Textbook

01:28

Evaluate the given integral.$$\int e^{a x} d x$$

01:54

Use the method of Lagrange multipliers to optimize $f$ as indicated, subject…

01:58

Evaluate the given integral.$$\int_{0}^{1} \sqrt{5 x+4} d x$$

02:20

Evaluate the given integral.$$\int \frac{3 x+2}{\left(3 x^{2}+4 x+1\righ…

Evaluate the given definite integral.(a) $\int_{2}^{3} 3 x d x$

06:19

Evaluate the given integral.$$\int_{0}^{4} \int_{x}^{2 x}\left(x^{2}+3 y…

03:08

Show that the tangent line to any point on the circle $x^{2}+y^{2}=a^{2}$ is…

05:23

Find the partial derivatives with respect to (a) $x$ and (b) $y$.$$f(x, …

00:43

$f(x, y)=4 x^{3} y^{2}$ determine (a) $f(3,2),(b) f(2,5)$

02:54

Use the properties about odd and even functions to evaluate the given integr…