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Find and classify, using the second partial derivative test, the critical points of the function defined .$$f(x, y)=4 x^{2}-2 y^{2}+16 x-20 y-15$$

$$\operatorname{sp}(-2,-5,19)$$

Calculus 3

Chapter 6

An Introduction to Functions of Several Variables

Section 3

Extrema

Partial Derivatives

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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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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 four, X squared minus two, Y squared plus 16 X minus 20 y minus 15. So to begin, we want to find where the gradient of F is going to equal zero. Now the first element, the partial derivative with respect to X is going to be 16 plus eight X. The second element is going to be negative 20 minus four. Y. We want that to equal the zero vector The solution to that is going to be x equals -2. Why equals -5. Now having that, that tells us that are critical point is going to be negative to negative five. Now we want to actually get on to the test which will require taking the second partial greatness. So we can see that the second partial derivative with respect to X is going to be eight. The second partial derivative with respect to Y is going to be negative four And the mixed partial is going to be equal to zero. So now that means that our d function for classification is going to come out to -32. Clearly that is less than zero. And since that D is less than zero, that tells us that the point negative to negative five is a saddle point

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