Problem 4 (7 points). Find all critical points of f(x, y) =...

Problem 4 (7 points). Find all critical points of $f(x, y) = x^2 + y^4 + 2xy$ and determine (to the best of your ability) whether each one is a local maximum, local minimum, a saddle point, or none of the above.

Transcript

0:00 Either.

00:02 In this problem we're asked to find all the critical points for this function and then figure out whether each is a local min, max, saddle point, or none of those things.

00:14 So to find critical points, we take the partial derivatives with respect to x and y, set them equal to zero and solve.

00:24 So we always need to start by taking the partial derivatives.

00:28 So let's take the partial derivative with respect to x and that will give us some 2x here.

00:36 The derivative of 4x with respect to x is 4.

00:40 The derivative of y squared with respect to x is just 0.

00:46 And the partial with respect to y, going back up to the top here, x squared doesn't have a y, so that's 0.

00:55 4x doesn't have a y, so that's 0.

00:58 Y squared has a partial derivative 2y.

01:04 Okay.

01:07 Now we want to set each of these equal to 0.

01:10 And then solve.

01:11 And we can do this quickly by inspection.

01:14 This first equation, the only way 2x plus 4 equals 0 is if x equals negative 2.

01:24 And the only way 2 y equals 0 is if y equals 0.

01:29 So we only have one critical point.

01:34 We have critical point, and it's at the point negative 2, comma 0.

01:42 Ok, so now all that's left is to determine whether we have local min, local max, saddle point, or neither of those.

01:50 So let's use the criterion right at the end of the chapter, the second derivative test as it's called.

02:01 So it involves the second partial derivatives...

Question

Problem 4 (7 points). Find all critical points of $f(x, y) = x^2 + y^4 + 2xy$ and determine (to the best of your ability) whether each one is a local maximum, local minimum, a saddle point, or none of the above.

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