Find the relative maxima or minima in Exercises 1-10 Minimum of f(x, y)=4 x^2+y^2-4 x y, subject

Find the relative maxima or minima in Exercises 1-10 Minimum...

Key Concepts

Constrained Optimization

A branch of optimization where the objective is to find the maximum or minimum of a function subject to certain restrictions or constraints. This is essential when the variables involved are related by one or more equations, and it requires methods that account for these dependencies.

Lagrange Multipliers

A mathematical strategy used to solve constrained optimization problems. By introducing additional variables (the Lagrange multipliers), the constraint is incorporated into the objective function, making it possible to locate extrema by finding the stationary points of the resulting Lagrangian function.

Quadratic Forms

Expressions involving variables where each term is of degree two. In optimization, quadratic functions often exhibit distinct curvature properties that help identify whether a critical point corresponds to a minimum, maximum, or saddle point, making them a common focus in problems involving extrema.

Transcript

00:01 Okay, so for this question, our function, f of x, y, that we're trying to optimize, is for x squared plus y squared minus 4xy, and the constraint is that x plus y equals 9.

00:18 So that gives us our constraint function, our g of x, y, to be x plus y minus 9.

00:29 Then the technique that we're going to use is we're going to create another function of three variables, f of x, y, gamma, where gamma is the garage multiplier, and that is f of x, y, minus gamma times g of xy, and we get 4x squared plus y squared minus 4xy minus 4xy, minus 4xy, minus 4xy, minus gamma x minus gamma y plus gamma times nine be sure that you're distributing this negative to all of those pieces that's a common mistake and then the technique that we're using says that once you set up this equation this new equation then we set up a corresponding system of three equations with three unknowns where we take the partial derivative with respect to x and i'm going to leave off the the independent variables there.

01:36 So partial derivative of this with respect to x is 8x minus 4y minus lambda equals 0.

01:46 So that's our first equation.

01:48 F of y is going to be 2y minus 4x minus lambda equal 0.

02:01 That's our second equation.

02:04 And then our third equation is f of lambda, so partial derivative with respect to lambda, which is minus x minus y plus nine, equals zero.

02:15 And then we use these three equations to figure out x and y and lambda.

02:22 Now there are several ways of solving the system.

02:26 I'll show you one way, so we can use equation one, solve that for x to get that x equals lambda plus 4y over 8, and then let's call this equation 4.

02:42 And then we can use equations 2 and 4.

02:44 So what we're going to do is we're just going to substitute this in for x there.

02:49 So we get 2y plus minus 4 times lambda plus 4 y over 8, minus lambda equals 0.

03:06 Doing some algebra, we get lambda.

03:09 Plus 4y over 2 minus lambda equals 0...

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