# MTH5P1: Maximum and minimum points

##### This unit explains Maximum and minimum points

Gradient at a maximum or minimum point

Figure 5.14 shows the graph of It has a maximum point at (0, 16). You will see that
● at the maximum point the gradient is zero

● the gradient is positive to the left of the maximum and negative to the right of it.
This is true for any maximum point (see figure 5.15). In the same way, for any minimum point (see figure 5.16):
● the gradient is zero at the minimum
● the gradient goes from negative to zero to positive. Maximum and minimum points are also known as stationary points as the gradient function is zero and so is neither increasing nor decreasing.

Find the stationary points on the curve of and sketch the curve.

SOLUTION
The gradient function for this curve is The signs of the gradient function just either side of these values tell you the nature of each stationary point. Thus the stationary point at x = −1 is a maximum and the one at x = 1 is a minimum.
Substituting the x values of the stationary points into the original equation, There is a maximum at (−1, 3) and a minimum at (1, −1). The sketch can now be drawn (see figure 5.19). In this case you knew the general shape of the cubic curve and the positions of all of the maximum and minimum points, so it was easy to select values of x for which to test the sign of The curve of a more complicated function may have several maxima and minima close together, and even some points at which the gradient is undefined. To decide in such cases whether a particular stationary point is a maximum or a minimum, you must look at points which are just either side of it.

Find all the stationary points on the curve of and sketch the curve.

SOLUTION You may find it helpful to summarise your working in a table like the one below.
You can find the various signs, + or −, by taking a test point in each interval, for  example t = 0.25 in the interval  There is a maximum point when t = 0 and there are minimum points when t = −0.5 and +0.5. Therefore (0, 1) is a maximum point and (−0.5, 0.875) and (0.5, 0.875) are minima.

The graph of this function is shown in figure 5.20. Increasing and decreasing functions

When the gradient is positive, the function is described as an increasing function. Similarly, when the gradient is negative, it is a decreasing function. These terms are often used for functions that are increasing or decreasing for all values of x. SOLUTION   Points of inflection This is an example of a point of inflection. In general, a point of inflection occurs where the tangent to a curve crosses the curve. This can happen also when as shown in figure 5.24. If you are a driver you may find it helpful to think of a point of inflection as the point at which you change from left lock to right lock, or vice versa. Another way of thinking about a point of inflection is to view the curve from one side and see it as the point where the curve changes from being concave to convex.

The second derivative

Figure 5.25 shows a sketch of a function y = f(x), and beneath it a sketch of the corresponding gradient function  Sketch the graph of the gradient of against x for the function illustrated in figure 5.25. Do this by tracing the two graphs shown in figure 5.25, and extending the dashed lines downwards on to a third set of axes.
You can see that P is a maximum point and Q is a minimum point. What can you say about the gradient of at these points: is it positive, negative or zero?

The gradient of any point on the curve of is given by This is written and is called the second derivative. It is found by differentiating
the function a second time.  Using the second derivative

You can use the second derivative to identify the nature of a stationary point, instead of looking at the sign of just either side of it.

Stationary points

Notice that at P, This tells you that the gradient, is zero and decreasing. It must be going from positive to negative, so P is a maximum point (see figure 5.26). This tells you that the gradient, is zero and increasing. It must be going from negative to positive, so Q is a minimum point (see figure 5.27). The next example illustrates the use of the second derivative to identify the nature of stationary points.        Applications

There are many situations in which you need to find the maximum or minimum value of an expression. The examples which follow, and those in Exercise 5G, illustrate a few of these.
EXAMPLE 5.16 Kelly’s father has agreed to let her have part of his garden as a vegetable plot. He says that she can have a rectangular plot with one side against an old wall. He hands her a piece of rope 5 m long, and invites her to mark out the part she wants. Kelly wants to enclose the largest area possible. What dimensions would you advise her to use?
SOLUTION
Let the dimensions of the bed be x m × y m as shown in figure 5.32.    The chain rule Your first thought may be to write it as and then get rid of the brackets, but that is not possible in this case because the power is not a positive integer.

Instead you need to think of the expression as a composite function, a ‘function of a function’.
You have already met composite functions in Chapter 4, using the notation g[f(x)] or gf(x).
In this chapter we call the first function to be applied u(x), or just u, rather than f(x). This is now in a form which you can differentiate using the chain rule.
Differentiating a composite function

To find for a function of a function, you consider the effect of a small change in x on the two variables, y and u, as follows. A small change δx in x leads to a small change δu in u and a corresponding small change δy in y, and by simple algebra,  SOLUTION
As you saw earlier, you can break down this expression as follows. Notice that the answer must be given in terms of the same variables as the question, in this case x and y. The variable u was your invention and so should not appear in the answer.

You can see that effectively you have made a substitution, in this case This transformed the problem into one that could easily be solved.
Note
Notice that the substitution gave you two functions that you could differentiate. Some substitutions would not have worked. For example, the substitution would give you You would still not be able to differentiate y, so you would have gained nothing. A student does this question by first multiplying out to get a polynomial of order 8. Prove that this heavy-handed method gives the same result.

With practice you may find that you can do some stages of questions like this in your head, and just write down the answer. If you have any doubt, however, you should write down the full method.

Differentiation with respect to different variables
The chain rule makes it possible to differentiate with respect to a variable which does not feature in the original expression. For example, the volume V of a sphere of radius r is given by Differentiating this with respect to r gives the rate of change of volume with radius However you might be more interested in finding the rate of change of volume with time, t. To find this, you would use the chain rule: You have now differentiated V with respect to t.
The use of the chain rule in this way widens the scope of differentiation and this means that you have to be careful how you describe the process.

‘Differentiate could mean differentiation with respect to x, or t, or any other variable. In this book, and others in this series, we have adopted the convention that, unless otherwise stated, differentiation is with respect to the
variable on the right-hand side of the expression. So when we write ‘differentiate or simply ‘differentiate it is to be understood that the differentiation is with respect to x.

The expression ‘increasing at a rate of’ is generally understood to imply differentiation with respect to time, t.

The radius r cm of a circular ripple made by dropping a stone into a pond is increasing at a rate of At what rate is the area enclosed by the ripple increasing when the radius is 25 cm? ## STEM Elearning

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