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Find and classify, using the second partial derivative test, the critical points of the function defined .$$f(x, y)=\frac{8}{x}+\frac{1}{2 y}+2 x y$$

$$\mathrm{rm}(4,1 / 4,6)$$

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 eight over X plus 1/2 Y plus two times X Y. Now to begin we want to find X and Y. Such that the gradient of F equals the zero factor. So the first element of the gradient of F. It's going to be two Y -8 over x squared. The second element is going to be two x minus 1/2, Y squared. Both of those are from taking partial derivatives with respect to X and Y. And we want those two equals zero and zero. The solution to that system of two equations is going to be X equals four, Y equals 1/4. Giving us our critical point. Now we want to proceed forward with doing our test. Now the second partial derivative with respect to X is going to be 16 over X cube. The second partial derivative with respect to why is going to be won over white cube. And the mixed partial derivative is going to be too. So our D r d function of X. Y which will be using for classification is going to be equal to 16 over x cubed y cube. Now we can or 16 over executed Y cube minus four. Now we can see that X cubed times y cube since X and y are in verses of each other. That that will multiply out to once you will get 16 minus four or just 12 greater than zero. And we can see that the second partial derivative with respect to X, is greater than zero, that's zero. Putting those two facts together. We can see that we will have a relative minimum at the point four,

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