Lagrange Multipliers

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Lagrange Multipliers

Based on “Lagrange Multipliers” in section 3.8 of the textbook.

Key Ideas

The notions of an objective function and constraint (see Tuesday’s notes or the textbook for definitions).

The equation relating the gradients of the objective and constraint functions (see the examples below, Tuesday’s notes, or the book for this equation).

The textbook’s 4-step strategy for using the equation.

The derivation of the equation, i.e., the fact that it comes from somewhere.

Examples

Finish the example we started yesterday, namely finding the extreme values of z = 2x² + 3y² - 4x - 6y + 5 over the ellipse x² + 2y² = 9. Yesterday we set up the basic equations from the method of Lagrange multipliers; now we should try to solve them.

To start, break the gradient equation into separate equations for each component, showing that those equations along with the constraint do at least give us as many equations to work with as there are variables:

This is not a system of linear equations, and that’s common with Lagrange multipliers. You’ll have to use a bit of creative problem solving to solve the system, you can’t rely on a general scheme that will always work. In this case, using the first 2 equations to express everything in terms of one variable, and then plugging that into the constraint and solving seems like a promising overall plan. The easiest variable to express everything else in terms of turns out to be λ:

At the end of this process, we divided by 4 - 2λ and 6 - 4λ. Whenever dividing by something, you need to be sure it’s not 0. That will be the case for these divisions as long as λ isn’t 2 or 3/2. But we left notes to remind ourselves to come back and check that there aren’t solutions based on those values, which we’d miss by doing the division, i.e., that we didn’t “divide away solutions.”

Plugging the expressions for x and y in terms of λ into the constraint equation gives an equation that we can in principle solve for λ:

But this is an ugly equation, and there’s no particular learning goal in solving it by hand, so we turned to Mathematica for that. You can download the notebook we created to see what we did or to experiment with variations.

Mathematica gave us x and y values at which the z reaches its minimum and maximum around the ellipse. But there’s one thing left to do: make sure that there aren’t other extremes corresponding to the values of λ that we “divided away” earlier. In both cases those values are λ turn out to be impossible: λ = 2 would make the first of the gradient equations unsolvable, and λ = 3/2 would make the second unsolvable. So we don’t have to worry about the division we did, but it’s important that we checked.

Another example: what is the maximum value of a “Cobb-Douglas function” p(x, y) = 2 x^0.4 y^0.6 with the constraint that x + y = 10?

We had time to set up the equations needed to use Lagrange multipliers, but not to solve them:

These equations will be easier to solve by hand than the first example was. See what you can do with them between now and Friday. We’ll start Friday’s class by finishing them.

Next

Finish the Cobb-Douglas example of Lagrange multipliers.

Then start talking about integrals of multivariable functions.

We’ll do the main reading about integrals for Monday, but if you want to get a head start, see “Volumes and Double Integrals,” “Properties of Double Integrals,” and “Iterated Integrals” in section 4.1 of the textbook, and “Integrable Functions of Three Variables” in section 4.4.

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