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Find and classify, using the second partial derivative test, the critical points of the function defined .$$f(x, y)=6 x y+4 y-\ln x y^{2}$$

$$\mathrm{rm}(2 / 3,1 / 4,2+\ln 24)$$

Calculus 3

Chapter 6

An Introduction to Functions of Several Variables

Section 3

Extrema

Partial Derivatives

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Lectures

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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. Six X. Y plus four Y minus lawn of X. Y squared. To do this, we first want to find the critical points by solving for when the gradient of F will equal zero. So the first element of the gradient is going to be six Y -1 over X. The second element is going to be six X plus four minus two over Why? Now if we want to solve for when this equals zero we have one solution. That being when X equals 2/3 And y equals 1/4. Having our critical point. Now Proceed with a flying our 2nd partial derivative test. So to do that we need the second partial X is going to be one over X squared. You need to be second parcel Why? Which is going to be two overlays squared. And we need the mixed partial Which is going to equal six than our D. Of X. Y function is going to equal to over X squared Y squared minus 36. And plugging in the value of our critical point there to over 3 1/4 We get a final value of positive 36. So that's greater than zero. We can see that the second partial derivative with respect to X will be greater than zero there as well, which tells us that the point 2/3 1/4 will be a relative minimum

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