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Find and classify, using the second partial derivative test, the critical points of the function defined .$$f(x, y)=x y+\frac{8}{y}+\frac{8}{x}$$

$$\mathrm{rm}(2,2,12)$$

Calculus 3

Chapter 6

An Introduction to Functions of Several Variables

Section 3

Extrema

Partial Derivatives

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University of Michigan - Ann Arbor

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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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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 x. Y plus eight over Y plus eight over X. So the first step here is to find where the gradient of our function is going to equal zero. Now the first element of our gradient is going to be y minus eight over X squared from the partial derivative with respect to X. The second element is going to be x minus eight over y squared Partial derivative. With respect why we want that to equal 00. Now we get the solutions that either X equals two and y equals two or X equals actually no, I'm going to take that back. We only really get one meaningful solution. X equals two lights too. Having that. Then we move on to actually testing, We take the second partial derivative with respect to X which is going to give us 16 over execute we take the second partial derivative with respect to why? Which is going to be 16 over Y. Cube. And we take our mixed partial derivative which is just going to equal one. So we then have our D of X. Y function which will be using for determining the nature of the critical point. It's going to come out to 2 56 X cube or 2 56 over x cubed y cube minus one. Now plugging in our .22 um one moment here. So 2 56/2 cubes times to cuba is going to come out to four. So we get four minus one equals three greater than zero. And we can see that the second partial derivative with respect to X at the 00.22, is going to be greater than zero as well, which tells us by our second partial derivative test that we will have a relative minimum at the point 22.

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