Now we put our optimization skills to work.

**optimization problem**is a problem where you need to maximize or minimize some quantity given some constraints. This can be accomplished using the tools of differential calculus that we have already developed.

Perhaps the most basic optimization problems is generated by the following question:

Among all rectangles of a fixed perimeter, which has the greatest area?

Let’s not do this problem in the abstract, let’s do it with numbers.

So to maximize the area of the rectangle, we need to find the maximum value of on the appropriate interval.

**At this point, you should graph the function if you can.**

We’ll continue on without the aid of a graph, and use the derivative. Write Now we find the critical points, solving the equation we see that the only critical point of is at

Since is positivenegative on and positivenegative on , is where the maximum value of happens. This is exactly when the rectangle is a square!

A key step to note, is when we explained why is actually the maximum. Above we basically used facts about the derivative. Below we use a similar argument.

If denotes one of the sides of the rectangle, then the adjacent side must be .

The perimeter of this rectangle is given by We wish to minimize . Note, not all values of make sense in this problem: lengths of sides of rectangles must be positive, so . If then so is , so we need no second condition on .

**At this point, you should graph the function if you can.**

We next find and set it equal to zero. Write Solving for gives us . We are interested only in , so only the value is of interest. Since is defined everywhere on the interval , there are no more critical values, and there are no endpoints. Is there a local maximum, minimum, or neither at ? The second derivative is and , so there is a local minimum. Since there is only one critical point, this is also the global minimum, so the rectangle with smallest perimeter is the square.

Hence, calculus gives a **reason** for **why** a square is the rectangle with both

- the largest area for a given perimeter.
- the smallest perimeter for a given area.

We may be done with rectangles, but they aren’t done with us. Here is a problem where there are more constraints on the possible side lengths of the rectangle.

**At this point, you should graph the function if you can.**

Setting we find as the only critical point. Testing this and the two endpoints (as the maximum could also be there), we have and . Hence, the maximum area occurs when the rectangle has dimensions .

Again, note that above we used the Extreme Value Theorem to guarantee that we found the maximum.