Maxima and Minima

One of the principal uses of the process of differentiating is to find out under what conditions the value of the thing differentiated becomes a maximum, or a minimum. This is often exceedingly important in engineering questions, where it is most desirable to know what conditions will make the cost of working a minimum, or will make the efficiency a maximum.

Now, to begin with a concrete case, let us take the equation

By assigning a number of successive values to , and finding the corresponding values of , we can readily see that the equation represents a curve with a minimum.

These values are plotted in the following figure, which shows that has apparently a minimum value of , when  is made equal to . But are you sure that the minimum occurs at , and not at or at ?

Of course it would be possible with any algebraic expression to work out a lot of values, and in this way arrive gradually at the particular value that may be a maximum or a minimum.

Here is another example:

Let .

Calculate a few values thus:

Plot these values as in the following figure.

It will be evident that there will be a maximum somewhere between and ; and the thing looks as if the maximum value of  ought to be about . Try some intermediate values. If , ; if , ; if , . How can we be sure that is the real maximum, or that it occurs exactly when ?

Now it may sound like juggling to be assured that there is a way by which one can arrive straight at a maximum (or minimum) value without making a lot of preliminary trials or guesses. And that way depends on differentiating. Look back to the previous chapter for the remarks about Fig. 10.8 and Fig. 10.9, and you will see that whenever a curve gets either to its maximum or to its minimum height, at that point its . Now this gives us the clue to the dodge that is wanted. When there is put before you an equation, and you want to find that value of  that will make its  a minimum (or a maximum), first differentiate it, and having done so, write its as equal to zero, and then solve for . Put this particular value of  into the original equation, and you will then get the required value of . This process is commonly called “equating to zero.”

To see how simply it works, take the example with which this chapter opens.

Try the second example (Fig. 11.2).

Now, before we go on to any further cases, we have two remarks to make. When you are told to equate to zero, you feel at first (that is if you have any wits of your own) a kind of resentment, because you know that has all sorts of different values at different parts of the curve, according to whether it is sloping up or down. So, when you are suddenly told to write you resent it, and feel inclined to say that it can’t be true. Now you will have to understand the essential difference between “an equation,” and “an equation of condition.” Ordinarily you are dealing with equations that are true in themselves, but, on occasions, of which the present are examples, you have to write down equations that are not necessarily true, but are only true if certain conditions are to be fulfilled; and you write them down in order, by solving them, to find the conditions which make them true. Now we want to find the particular value that has when the curve is neither sloping up nor sloping down, that is, at the particular place where . So, writing does not mean that it always is ; but you write it down as a condition in order to see how much will come out if is to be zero.

The second remark is one which (if you have any wits of your own) you will probably have already made: namely, that this much-belauded process of equating to zero entirely fails to tell you whether the  that you thereby find is going to give you a maximum value of  or a minimum value of . Quite so. It does not of itself discriminate; it finds for you the right value of  but leaves you to find out for yourselves whether the corresponding  is a maximum or a minimum. Of course, if you have plotted the curve, you know already which it will be.

For instance, take the equation:

Without stopping to think what curve it corresponds to, differentiate it, and equate to zero: whence and, inserting this value,

will be either a maximum or else a minimum. But which? You will hereafter be told a way, depending upon a second differentiation (see Chapter [Ch:Curvature]). At the end of this chapter, an alternative method will be presented, which explains how to distinguish between a maximum and a minimum based on the sign of the derivative. But at present it is enough if you will simply try any other value of  differing a little from the one found, and see whether with this altered value the corresponding value of  is less or greater than that already found.

Try another simple problem in maxima and minima. Suppose you were asked to divide any number into two parts, such that the product was a maximum? How would you set about it if you did not know the trick of equating to zero? I suppose you could worry it out by the rule of try, try, try again. Let be the number. You can try cutting it into two parts, and multiplying them together. Thus, times is ; times is ; times is ; times is ; times is . This looks like a maximum: try varying it. times is , which is not so good; and times is , which is worse. So it seems that the biggest product will be got by dividing into two equal halves.

Now see what the calculus tells you. Let the number to be cut into two parts be called . Then if is one part, the other will be , and the product will be or . So we write . Now differentiate and equate to zero; Solving for , we get So now we know that whatever number may be, we must divide it into two equal parts if the product of the parts is to be a maximum; and the value of that maximum product will always be .

This is a very useful rule, and applies to any number of factors, so that if a constant number, is a maximum when .

Test Case

Let us at once apply our knowledge to a case that we can test.

Example 11.3. Let and let us find whether this function has a maximum or minimum; and if so, test whether it is a maximum or a minimum.

Solution. Differentiating, we get Equating to zero, we get whence or

That is to say, when is made , the corresponding value of  will be either a maximum or a minimum. Accordingly, putting in the original equation, we get or

Is this a maximum or a minimum? To test it, try putting a little bigger than ,—say make . Then which is higher up than ; showing that is a minimum.

The curve is plotted below. As we can see, the minimum of is , which occurs when .

Further Examples

A most interesting example is afforded by a curve that has both a maximum and a minimum.

Example 11.4. Consider a curve whose equation is: Now

Equating to zero, we get the quadratic, and solving the quadratic gives us two roots, viz.

Now, when , ; and when , . The first of these is a minimum, the second a maximum.

The curve itself may be plotted (as in the following figure) from the values calculated, as below, from the original equation.

					1

A further exercise in maxima and minima is afforded by the following example:

Example 11.5. The equation to a circle of radius  (), having its centre  at the point whose coordinates are , , as depicted below, is:

This may be transformed into

Now we know beforehand, by mere inspection of the figure, that when ,  will be either at its maximum value, , or else at its minimum value, . But let us not take advantage of this knowledge; let us set about finding what value of  will make a maximum or a minimum, by the process of differentiating and equating to zero. which reduces to

Then the condition for being maximum or minimum is:

Since no value whatever of  will make the denominator infinite, the only condition to give zero is Inserting this value in the original equation for the circle, we find and as , we have two resulting values of ,

The first of these is the maximum, at the top; the second the minimum, at the bottom.

If the curve is such that there is no place that is a maximum or minimum, the process of equating to zero will yield an impossible result. For instance:

Example 11.6. Let

Then

Equating this to zero, we get , and

which is impossible.¹ Therefore has no maximum nor minimum.

The following figure depicts the graph of with specific values for , , and . It is evident from the figure that the function does not possess any maximum or minimum points.

A few more worked examples will enable you to thoroughly master this most interesting and useful application of the calculus.

Example 11.7. What are the sides of the rectangle of maximum area inscribed in a circle of radius ?

Solution. If one side is called , and as the diagonal of the rectangle is necessarily a diameter, the other side  (see the following figure).

Then, area of rectangle ,

If you have forgotten how to differentiate , here is a hint: write and , and seek and ; fight it out, and only if you can’t get on refer to the Chain Rule on page .

You will get

For maximum or minimum we must have that is, and .

The other side ; the two sides are equal; the figure is a square the side of which is equal to the diagonal of the square constructed on the radius. In this case it is, of course, a maximum with which we are dealing.

Example 11.8. What is the radius of the opening of a conical vessel the sloping side of which has a length  when the capacity of the vessel is greatest?

.

Solution. If be the radius and the corresponding height, (see the following figure).

Proceeding as in the previous problem, we get for maximum or minimum.

Or, , and , for a maximum, obviously.

Example 11.9. Find the maxima and minima of the function Solution. We get for maximum or minimum; or

There is only one value, hence only one maximum or minimum. it is therefore a minimum.

The graph of is shown below.

Example 11.10. Find the maxima and minima of the function .

Solution. Differentiating gives at once (see example 9.1) for maximum or minimum.

Hence and , the only solution

For ,.

For , , so this is a maximum.

The graph of is shown below.

Example 11.11. Find the maxima and minima of the function

Solution. We have for maximum or minimum; or or ; which has for solutions

These being imaginary, there is no real value of  for which ; hence there is neither maximum nor minimum.

The following figure shows the graph of .

Example 11.12. Find the maxima and minima of the function

Solution. This may be written . that is, , which is satisfied for , and for , that is for . So there are two solutions.

Taking first . If , , and if , . On one side  is imaginary; that is, there is no value of  that can be represented by a graph; the latter is therefore entirely on the right side of the axis of  (see the following figure).

On plotting the graph it will be found that the curve goes to the origin, as if there were a minimum there; but instead of continuing beyond, as it should do for a minimum, it retraces its steps (forming what is called a “cusp”). There is no minimum, therefore, although the condition for a minimum is satisfied, namely . It is necessary therefore always to check by taking one value on either side.

Now, if we take . If , and ; if ,  becomes and ; and if ,  becomes and .

This shows that there are two branches of the curve; the upper one does not pass through a maximum, but the lower one does.

Make other examples for yourself. There are few subjects which offer such a wealth for interesting examples.

Exercises

Exercise 11.1. What values of  will make a maximum and a minimum, if ?

 

Answer

Min.: , ; max.: , .

 

 

Solution

 

When , .

When , .

When , .

Therefore is a minimum occurring when .

When , .

When , .

When , .

Therefore, is a maximum occurring when .

The graph of is shown below.

 

 

Exercise 11.2. What value of  will make a maximum in the equation ?

 

Answer

.

 

 

Solution

Then

 

To see which value of makes a maximum and which value of makes a minimum, it would be the best if we use the first derivative test:

If , then and hence If , then and hence

It follows from the First Derivative Test that correspondent to a maximum .

Similarly, if , then and hence If , then and hence . Therefore makes a minimum of .

We can rewrite the given equation as or

Now we can easily plot versus (see the following figure). As we can see, the graph reaches a maximum at and a minimum at . The maximum value of is and the minimum value of is .

 

 

Exercise 11.5. Plot the curve corresponding to the equation also find , and deduce the value of  that will make a minimum; and find that minimum value of .

 

Answer

; ; .

 

 

Solution

 

From the graph, it is clear that makes a minimum

When .