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 to identify the quantity that 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!

Although the function is continuous on its domain, its domain is an open, unbounded interval and the Extreme Value Theorem does not apply; there is no guarantee that the minimum even exists. But we don’t give up! Our strategy will be to search for a critical point and check if the minimum is attained there. By solving the equation we find that the point where is the only critical point of . Since is positive when is positive, there is a local minimum at the critical point, and hence a global minimum since there is only one critical point.

Another way to reach this conclusion is to examine the sign of the derivative on the interval : From this factorization it is clear that on , and on . This means that the function is decreasing on and increasing on , and, therefore, has a global minimum at the critical point.

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., , and . Note that is the maximum of these. Thus the maximum profit is $, attained when we set the price at $ and sell items.

Alternately, we could apply the Second Derivative Test at the critical point and conclude that there must be a local maximum there, since . Since this is the only critical point, it must be a global maximum as well. 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 between and 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
below.

Apply the Pythagorean Theorem to the right triangle in the figure above to obtain

Solve for , since is found in the formula for the volume of the cone, and find that Substitute this into the formula for the volume of the cone to find that By simplifying, we were able to express the volume of the cone as a function of a single variable, ,

We want to maximize the function on the interval . Since this is a continuous function on a closed interval, the maximum is attained either at a critical point or an end point. By solving the equation we find that the function has at its only critical point, since is an end point. We evaluate the function at all the critical and end points,

The maximum is obviously attained at . 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

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.**