Decide whether F=x^2 y^2-2 x-2 y has a minimum at the point x=y=1 (after showing that the first

Decide whether F=x^2 y^2-2 x-2 y has a minimum at the point...

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Key Concepts

Critical Points

Critical points of a function occur where the first derivatives (or the gradient) are zero or undefined. They are the candidate points where local minima, maxima, or saddle points can occur. Determining whether a function has a minimum at a critical point is a central task in multivariable optimization.

Gradient and Partial Derivatives

The gradient of a multivariable function is a vector consisting of its partial derivatives with respect to each variable. Evaluating the partial derivatives and setting them to zero helps in identifying the critical points of the function. This procedure is essential for solving optimization problems in higher dimensions.

Hessian Matrix and Second Derivative Test

The Hessian matrix is a square matrix of second-order partial derivatives of a function. It is used in the second derivative test for multivariable functions to classify the nature of a critical point. By analyzing the definiteness of the Hessian matrix at a critical point, one can determine whether the function has a local minimum, maximum, or saddle point at that position.

Transcript

00:01 Here we have a function, f of x, y, which is equal to x squared, y squared, minus 2x, minus 2y, and we want to determine whether it has a minimum at x equals y equals 1.

00:14 So the first step is to look at the first partial derivative.

00:19 So the first partial derivative with respect to x is going to be 2xy squared minus 2.

00:27 And when we evaluate that at 1 -1, what we're going to get is 2 minus 2 equals 0.

00:37 Okay, so potentially we have some sort of extremal point here.

00:45 Let's look at d, f, d, d, f, y.

00:47 Because of the symmetry in this equation, it's going to look very similar to x squared y minus 2, and then you can already guess what's going to happen when we plug in.

01:01 Sorry about that.

01:02 You can guess what's going to happen when we plug in 1 -1 because it's symmetric with the previous case.

01:07 We're going to get 0.

01:10 Okay, so now we want to check if we have a local minima, and to do that, we need to check the second, a local minimum, sorry.

01:19 So we have some sort of extremal point, or maybe a saddle point here.

01:25 Let's look at the second derivatives.

01:29 First, the d squared f of d.

01:32 D x squared.

01:34 So that's just going to be 2y squared...