We learn to optimize surfaces along and within given paths.

Below, we see a geometric interpretation of this theorem.

A similar theorem applies to functions of several variables.

When finding extrema of functions in your first calculus course, you had to check inside an interval and at the end-points. Now we need to check the interior of a region and the boundary curve. If we check those critical points, then the Extreme Value Theorem tells us that we have found the extrema! No other tests are necessary.

To check the interior of a region, one finds where the gradient vector is zero or
undefined—we did that in the last section. Hence in this section, we will focus on
how one finds extrema on the boundary. In this case, the Extreme Value
Theorem **still** applies, and so we will be sure that we found the extrema on
the boundary. Let’s see some examples, we start with a classic real-world
problem.

Given a rectangular box where the width and height are equal, what are the dimensions of the box that give the maximum volume subject to the constraint of the size of a Standard Post Package?

We now consider the volume along the constraint . Along this line, we have:

Set and find the critical points. Write with me: when and .

We found two critical values: when and when . We ignore the solution. Thus the maximum volume, subject to the constraint, comes at , . This gives a volume of .

This portion of the text is entitled “Constrained optimization” because we want to
find extrema of a function subject to a *constraint*, meaning there are limitations to
what values the function can attain. In our first example the constraint was set by
the U.S. Post office. Constrained optimization problems are an important topic in
applied mathematics. The techniques developed here are the basis for solving larger
problems, where the constraints are either more complex or more than two variables
are involved.

Since we are working within a closed and bounded set , we now find the maximum and minimum values that attains along . This means we find the extrema of along the edges of the triangle.

Start with the bottom edge of the triangle, along the line while runs from : Now set as this represents the bottom edge of the triangular boundary. We want to maximize/minimize on the interval . To do so, we evaluate at its critical points and at the endpoints. First find the critical points of : We see that is the only critical point of . Evaluating at its critical point and at the endpoints of gives: We need to do this process twice more, for the other two edges of the triangle. Along the left edge, along the line , we substitute in for in :

We want the maximum and minimum values of on the interval , so we evaluate at its critical points and the endpoints of the interval. First find the critical points of : We see that is the only critical point of . Evaluating at its critical point and the endpoints of gives: Finally, we evaluate along the right edge of the triangle, where :

We want the maximum and minimum values of on the interval , so we evaluate at its critical points and the endpoints of the interval. First find the critical points of : We see that is the only critical point. Evaluating at this critical point and at the endpoints of the interval : Check out the following graph:

When working constrained optimization problems, there is often more than one way to proceed. Let’s do the previous problem again, but this time we will use vector-valued functions and the chain rule.

We see that is the only critical point of . Evaluating this at its critical point and at the endpoints of gives:

We need to do this process twice more, for the other two edges of the triangle. For the left edge, write with me:

We see that is the only critical point of . Evaluating this at its critical point and at the endpoints of gives:

Now for the right edge, write with me:

We see that is the only critical point of . Evaluating this at its critical point and at the endpoints of gives:

Again, check out the following graph:

So far, our constraints have all been lines. This doesn’t have to be the case.

Now we set . Here you may be tempted to solve for , but this author suggestions that you instead solve for and . Note that if one solution is . This corresponds to the points Now assume that and divide the equation above by . We now have Using the quadratic formula we find This corresponded to points Plugging these points back into , we find that the minimum value of on is and the maximum value is .

It is hard to overemphasize the importance of optimization. As humans, we
routinely seek to make *something* better. By expressing the *something* as a
mathematical function, “making *something* better” means “optimize *some
function.*”