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Find and classify, using the second partial derivative test, the critical points of the function defined .$$f(x, y)=2 x^{2} y+3 x y^{2}-36 x y$$

$$\mathrm{TF}(0,0,0), \operatorname{sp}(18,0,0), \operatorname{sp}(0,12,0), \operatorname{sp}(6,4,-280)$$

Calculus 3

Chapter 6

An Introduction to Functions of Several Variables

Section 3

Extrema

Partial Derivatives

Johns Hopkins University

Harvey Mudd College

University of Michigan - Ann Arbor

Boston College

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.

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Find and classify, using t…

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01:51

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 two X squared y plus three X Y squared minus 36 X Y. To begin, we want to find the critical points by finding where the gradient of the function will equal zero. Now the first component of the ingredients is going to be three Y square. The second component is going to be four X. Y. Or excuse me, The first component is going to be three Y squared plus four X Y minus 36 Y. Then the second component is going to be six X Y plus two X squared minus 36 X. Now we want to set that equal to the zero vector. We'll find that there are a few possible solutions, particularly we'll have that X equals zero. And why even 12 is one solution. X equals six, Y equals four is another X equals 18, y equals zero is another and X equals zero, Y equals zero is another. The next step here is actually applying our test, which first of all we will have to start taking the second partial derivatives. Second partial X is going to be for y. Second partial Why? It's going to be six x. And the mixed partial is going to be uh four X plus six, Y minus 36. So our d function d of x Y is going to be equal to 24 X times Y minus negative 36 plus four X plus six Y. All squared. Now evaluating this at The .012 gives us negative 1296 Just less than zero. And we can see that the second partial derivative would be are actually Since we have the less than zero there, we can see that the .012 will be a saddle point. Then at the .64 We get a result of 432, Which clearly is greater than zero. And we can see that the second partial derivative with respect to X is going to be positive because we have a positive Y value, Which tells us then that they .64 is going to be a relative minimum than evaluating it at the 0.18 zero We get negative 1296 which indicates again that zero or not, 0 12, excuse me that 18 0 will be a saddle point And evaluating it at the zero. We again get a result of 1296, meaning 00 is a saddle point.

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