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

$$\mathrm{rm}(1,2,6)$$

Calculus 3

Chapter 6

An Introduction to Functions of Several Variables

Section 3

Extrema

Partial Derivatives

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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 two over X plus four over Y plus x times Y. So our first step is to find where the gradient of our function equals zero, gradient. The first element of the gradient is going to be y minus two over X squared from taking the partial derivative with respect to X. The second element is going to be x minus four over y squared from taking the partial derivative with respect to why now we want that to equal 00. Uh Which the solution to that that system of two equations is going to be X equals one. Y equals two. Having that point as our critical point candidate. Excuse me candidate. We now want to take the second partial derivatives. The second partial derivative with respect to X is going to be four over x cubed second partial derivative with respect to Y is going to be eight over y cube And the mixed partial derivative is going to equal one. Which means that our d function is going to come out to 32 over x cubed y cube minus one. So calculating that at the .1, 2 we'll have 32/1 times two cubed to Cuba is going to be eight. So we can see that without actually calculating that out specifically. That is clearly going to be greater than zero. And we can also see that the second the second partial derivative of respect to X will be greater than zero as well, which the second partial derivative test tells us that that means that we will have a relative minimum at the point 12.

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