When you are on top of a hill, you are at a local maximum although there may be other hills higher than the one on which you are standing. Similarly, when you are at the bottom of a valley, you are at a local minimum even though there may be other valleys deeper than the one you are in. The word, “local” is applied to the situation because if you confine your attention only to points close to your location, you are indeed at either the top or bottom.
Derivatives can be used to locate local maxima and local minima. The following picture suggests how to do this. This picture is of the graph of a function having a local maximum and the tangent line to it.
Note how the tangent line is horizontal. If you were not at a local maximum or local minimum, the function would be falling or climbing and the tangent line would not be horizontal.
Theorem 5.7.2 Suppose f : U → ℝ where U is an open subset of ℝ and suppose x ∈ U is a local maximum or minimum. Then f^{′}
Proof: Since U is an open set, there exists δ > 0 such that
Points at which the derivative of a function equals 0 are sometimes called critical points. Included in the set of critical points are those points where f^{′} fails to exist. You could end up with a local maximum or minimum at such a point. Think of y =
The following is a typical minimization problem, this one heavily dependent on geometry.
Example 5.7.3 Find the volume of the smallest right circular cone which can be circumscribed about a ball of radius 4 inches. Such a cone has volume equal to
Consider the following picture of a cross section in which l is the length of the line from the center of the ball to the to vertex of the cone as shown.
The angle between the indicated radius of length 4 and the side of the cone is π∕2 from geometric considerations. Thus l sinθ = 4 and the volume of the cone is

Then the volume of the cone is

It seems there should be a solution to this problem and so we only have to find it by taking a derivative and setting it equal to 0 because the solution will surely be a local minimum. To take the derivative, use the rules of differentiation developed above. The derivative is

Obviously you cannot have l = 4. Such a situation would not even give a triangle. Therefore, the solution to the problem involves

There are two solutions, l = 12 or l = −4, the latter making absolutely no sense at all. Hence l = 12 must be the answer and the height of the cone is 16. The minimum volume is then

I think you probably could not do this problem without the methods of calculus.
Now here is an example about minimizing cost. It is another example which you could not work without the methods of calculus.
Exercise 5.7.4 A cylindrical can is to have volume 20π cubic inches. The top costs 2 cents per square inch and the sides cost 1 cent per square inch. What is the radius of the can which costs as little as possible.
You need 20π = πr^{2}h and so r^{2}h = 20. Now the cost is

Then the total cost in terms of r is C = 40
This scheme in which you take the derivative and set it equal to zero might not find the answer. It only gives candidates for the answer on the interior of an interval. Perhaps, like the above two examples these are the only points of interest. However, in general, when you look for the absolute maximum or minimum, you must consider the end points of the interval also.
Example 5.7.5 Let f
There exists a maximum and a minimum by the extreme value theorem. These could occur on

Now x = 1,0,4 are all possibilities. f
The above illustrates how it is done in general. You consider critical points and end points. Then among these points, you find the one which gives the best answer.