Now we put our optimization skills to work.

- Draw a picture.
- If possible, draw a schematic picture with all the relevant information.
- Determine your goal.
- We need identify what needs to be optimized.
- Find constraints.
- What limitations are set on our optimization?
- Solve for a single variable.
- Now you should have a function to optimize.
- Use calculus to find the extreme values.
- Be sure to check your answer!

We want to know the minimum value of this function when is in . Setting we find . Since is positive when is positive, there is a local minimum at the critical value, and hence a global minimum since there is only one critical value.

Finally, since ,

so the minimum cost occurs when the height is times the radius. If, for example, there is no difference in the cost of materials, the height is twice the radius.

We want to know the maximum value of this function when is between 0 and . The derivative is which is zero when . Since , there must be a local maximum at , and since this is the only critical value it must be a global maximum as well. Alternately, we could compute , , and and note that is the maximum of these. Thus the maximum profit is $, attained when we set the price at $ and sell items. We can confirm our results by looking at the graph of :

Notice that the function we want to maximize, , depends on *two* variables. Our next
step is to find the relationship and use it to solve for one of the variables in terms of
the other, so as to have a function of only one variable to maximize. In this
problem, the condition is apparent in the figure, as the upper corner of the
triangle, whose coordinates are , must be on the circle of radius . Write
Solving for , since is found in the formula for the volume of the cone, we
find Substitute this into the formula for the volume of the cone to find

We want to maximize when is between and . We solve finding or . We compute and The maximum is the latter. Since the volume of the sphere is , the fraction of the sphere occupied by the cone is

The optimal solution likely has the line being run along the ground for a while, then underwater, as the figure implies. We need to label our unknown distances: the distance run along the ground and the distance run underwater. Recognizing that the underwater distance can be measured as the hypotenuse of a right triangle, we can label our figure as follows

Evaluating at gives a cost of about . The distance the power line is laid along land is and the underwater distance is . We can confirm out results by looking at the graph of :

We now work a similar problem without concrete numbers.

You travel the distance from to at speed , and then the distance from to at speed . The distance from to is . By the Pythagorean theorem, the distance from to is Hence the total time for the trip is We want to find the minimum value of when is between 0 and . As usual we set and solve for . Write We find that Notice that does not appear in the last expression, but is not irrelevant, since we are interested only in critical values that are in , and is either in this interval or not. If it is, we can use the second derivative to test it: Since this is always positive there is a local minimum at the critical point, and so it is a global minimum as well.

If the critical value is not in it is larger than . In this case the minimum must occur at one of the endpoints. We can compute

but it is difficult to determine which of these is smaller by direct comparison. If, as is likely in practice, we know the values of , , , and , then it is easy to determine this. With a little cleverness, however, we can determine the minimum in general. We have seen that is always positive, so the derivative is always increasing. We know that at the derivative is zero, so for values of less than that critical value, the derivative is negative. This means that , so the minimum occurs when .

So the upshot is this: If you start farther away from than then you always want to cut across the sand when you are a distance from point . If you start closer than this to , you should cut directly across the sand.

With optimization problems you will see a variety of situations that require you to
combine problem solving skills with calculus. Focus on the *process*. One must learn
how to form equations from situations that can be manipulated into what you need.
Forget memorizing how to do “this kind of problem” as opposed to “that kind of
problem.”

**Learning a process will benefit one far more than
memorizing a specific technique.**