Find the global maximum value and global minimum value of fon S and indicate where each occurs.

Find the global maximum value and global minimum value of...

Key Concepts

Completing the Square

Completing the square is a technique used to rewrite quadratic functions in a form that makes their properties, such as the vertex in the case of a parabola, more transparent. In the context of optimization, rewriting a quadratic function in its vertex form helps in identifying critical points and understanding the function's behavior.

Extreme Value Theorem

The Extreme Value Theorem guarantees that if a function is continuous on a closed and bounded (compact) set, then the function must attain both a global maximum and a global minimum on that set. This theorem provides the foundational assurance that problems involving optimization on such sets are well-posed.

Constrained Optimization

Constrained optimization deals with finding the maxima or minima of a function when the domain of the function is restricted by one or more constraints. It involves analyzing not only the behavior of the function in the interior of the domain but also along the boundary defined by the constraints.

Lagrange Multipliers

The method of Lagrange multipliers is a technique used to find the local extrema of a function subject to equality constraints. By equating the gradient of the function with a scalar multiple of the gradient of the constraint function, this method transforms a constrained problem into a system of equations that can be solved for potential extreme points.

Transcript

00:01 All right, for this problem, we are asked to find the global maximum value and global minimum value of the function f of xy equals x squared minus x squared minus ay plus 7 on the region s equals x and y such that x squared plus y squared is less than or equal to one, and to indicate where each occurs.

00:18 So to begin, we can try, excuse me, tripping over my words, we can try taking the partial derivatives of the function with respect to x and y, in which case we get 2x minus 6.

00:31 Partial with respect to y is going to be 2y minus 8, which would indicate that we'd have a critical value at x, y, equals 3, 4, which is not in our region s.

00:44 So that indicates that we won't have any critical points for our function, which then means that we'd have to be finding our global max and min along the boundary of our region.

00:55 So we have that not just x squared plus y squared is less than or equal to one, but instead we'd get that x squared plus y squared must equal one, which we can substitute into our f of x y.

01:08 First of all, that means that the x squared plus y squared will add up to one and we'll have that seven out front.

01:13 So we'll have an eight from that.

01:16 And then we can rearrange the x squared plus y squared equation to get, for instance, y squared equals one minus x squared.

01:22 So y equals the square root or one minus x squared to the power of one half.

01:28 So we get then that f of now just x, is going to equal, it was 8 minus 6x minus 8 times 1 minus x squared power of 1 half.

01:44 Let's see, daisy.

01:46 Now we want to figure out a way to minimize this function.

01:52 One second here.

01:54 So trying to minimize this new function with respect to just x, we take the partial derivative, or the now just singular derivative, we should get the result is 8x over the square root of 1 minus x squared minus 6, and then we want to set that to 0.

02:11 So we can then solve for x here.

02:16 What we'll do is first we can, yeah, first let's move that 6 over to the right -hand side and multiply both sides by root 1 minus x squared.

02:28 So we get 8x equals 6 times a square root of 1 minus x squared.

02:32 Then we can square both sides.

02:35 So we'll have 8 squared, i believe, is 64 x squared, equals 36 times 1 minus x squared...

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