Find all the critical points and determine whether each is a local maximum, local minimum, a sad

Find all the critical points and determine whether each is a...

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Key Concepts

Second Derivative Test

This test uses the Hessian matrix to classify critical points based on the curvature of the function. By evaluating the determinant and the signs of the second-order partial derivatives at a critical point, one can determine whether the point is a local minimum, local maximum, or a saddle point.

Hessian Matrix

The Hessian matrix is a square matrix of second-order partial derivatives of a function. It provides information on the local curvature of the function at a critical point, and is crucial in determining the nature of that critical point in terms of local minima, maxima, or saddle points.

Gradient

The gradient of a function is a vector that comprises all its first partial derivatives. It indicates the direction of the steepest ascent and its magnitude gives the rate of increase, making it a fundamental tool for identifying critical points where the gradient equals zero.

Critical Points

These are points in the domain of a function where its gradient is either zero or undefined. In the context of multivariable calculus, finding critical points is essential as they are candidates for local extrema or saddle points, which are the points of interest in optimization problems.

Transcript

0:00 Hi there.

00:01 In this problem, we're asked to find any critical points here and then figure out whether we have local men, max, saddle point, or neither.

00:09 So whenever we're finding critical points, the very first thing we do is take the partial derivatives with respect to x and y because we want to set them equal to zero.

00:17 So let's do that now.

00:19 Partial with respect to x.

00:22 Now x squared becomes 2x.

00:25 X, y, remember we're treating y like it's a constant and x like the variable.

00:30 So the partial derivative with respect to x becomes y, and 3y doesn't even have an x in it, so that becomes 0.

00:38 Partial derivative with respect to y, let's go back to the top.

00:43 X squared is 0.

00:46 X, y, that partial derivative with respect to y, is just x, and the partial derivative of 3y with respect to y is 3.

00:55 So we have our partials.

00:57 The whole point of that is so that we can set these equal to 0 to find our critical point.

01:03 Solve each of these equations is easiest to solve this bottom equation because the only way x plus 3 could equal 0 is if x equals minus 3.

01:19 And now, plugging that into the top equation, if x must equal minus 3, then we have 2 times minus 3 plus y equals 0.

01:33 In other words, negative 6 plus y equals 0, so why would have to equal 6...

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