Absolute Extreme Values

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Problem Set

…On directional derivatives, gradients, extreme values.

Work on it this week, grade it next.

See the handout for details.

Local Extrema and the Second Derivative Test

Finishing Friday’s discussion of local extrema.

Friday we found that the function f(x,y) = cos x + sin y has critical points at (0, π/2) and (0, -π/2).

Use the second derivative test to verify that the critical point at (0, π/2) corresponds to a local maximum, and the critical point at (0, -π/2) to a saddle point.

Start by finding the first and second partial derivatives of f:

Then plug them into the discriminant formula, and interpret the signs of the results according to the rules of the test (see the textbook or Friday’s notes for those rules). Note that when the discriminant is positive, it means you have a maximum or minimum, but you need to additionally look at second derivatives to know which.

Absolute Extreme Values

Based on “Absolute Maxima and Minima” in section 3.7 of the textbook.

Questions

Where does the complexity of the book’s examples come from?

The problem is that Fermat’s Theorem (the theorem that says extreme values happen at critical points) requires the function in question to be defined on an open set around the critical point of interest. If you’re trying to find minimums or maximums over some region, that “open set” requirement isn’t a problem inside the region. But right on the boundaries of the region, if they’re included in it (i.e., if the region is closed), then you have a problem: points on the boundary don’t have open sets surrounding them that are entirely inside your region. So you have to look for maximums and minimums on the boundaries of your region separately from finding critical points. This is exactly the same in theory as separately checking the endpoints of a closed interval in single-variable calculus, but regions in 2 or more dimensions can have much more complicated boundaries than intervals in 1 dimension do, thus the added complexity in checking them.

Key Ideas

f(x,y,…) has an absolute minimum and maximum over a closed region.

Strategy for finding them:

1. Find critical points inside the region and evaluate the function at them
2. Find minimum and maximum values on the boundary of the region
3. The absolute minimum is the minimum of the function values from (1) and (2); the absolute maximum is the maximum of the function values.

Example

Find the absolute minimum and maximum for the function z = 2x² + 3y² on the triangle in the xy-plane with corners (1,0), (-1,-1), and (-1,1).

We started by finding critical points (there was one at the origin) and noting that we’ll need the function’s values at the corners of the triangle.

Now we also need to know if there are minimums or maximums along the boundary of the triangle. To do this, we found an equation for one of the sides (I caught a mistake in the equation as given in class, the Y intercept should be positive. That’s fixed here):

Think about how this equation (and others for the other sides of the triangle) might help. Take a look at the book and its examples to see if you can find it working with anything similar. Then we’ll talk about these things and finish checking for absolute minima and maxima tomorrow.

Next

Finish absolute extreme values.

Start constrained optimization in general terms.

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