chap14e.html

Chapter 14: Maximum/Minimum Problems

A Procedure for solving max/min story problems.

A max/min problem is a story problem where we want to optimize something, that is, we want to determine the situation where something is the biggest possible (such as profit, volume of a container, etc.) or where something is the smallest possible (cost, amount of material needed, time of travel, etc.) Solving a max/min problem is the two-step process of:

(1) reformulating the story problem as a ``pure math problem,'' and

(2) solving the ``pure math problem''.

We begin with a brief outline of the procedure, then discuss how to solve the pure math problem, and then present several examples illustrating the procedure.

The pure math problem just alluded to was described at the beginning of Chapter 13:

Problem: Find the maximum (or minimum) value of a function f(x) on the interval from a to b.

Here is a procedure for solving this type of problem:

(1) Find the value of f(x) at each point x between a and b where
f '(x) = 0.

(2) Find the value of f (x) at all points x between a and b where f '(x) does not exist.

(3) Find the value of f(x) at the endpoints a and b. (If and/or you have to find the values that f(x) approaches as x approaches +/- .

(4) The maximum (minimum) value of f(x) is the largest (smallest) of the
values found in steps (1)--(3).

Here are three simple examples which illustrate the procedure.

Worked examples of max/min problems

Example 1 : Let . Find the maximum value of x on the interval between 0 and 4.

First note that = 0 when , and . Next observe that is defined for all , so step (2) gives no x-values. Finally, and . The largest value of f(x) on the interval between 0 and 4 is f .

Example 2: Again let f(x) = - . Find the maximum value of on the interval between and .

First note that = when . But is not in the interval, so we don't use it. Next observe that is defined for all , so step (2) gives no -values. Finally, and . So the largest value of f(x) on the interval between -1 and 1 is .

Note: The maximum value depends not only on the function but on the interval as well. Later in this chapter, we will give examples of story problems where this observation is important.

Example 3: Find the minimum value of the function for
between and .

Note that for and for .
Thus the function f(x) is given by two separate formulas: f(x) = 7+2-x = 9-x

for x less than 2 and f(x) = 7+x-2 = 5+x for x greater than or equal to 3 and less than or equal to 4.

code for piecewise expressions

Thus the derivative of f(x) is given by:

[Maple Math] and

The derivative is not defined at the end points, is also undefined and is never zero . So we compute . Finally, we check the end points and f(4) = 9. The smallest of these numbers is , which is, therefore, the minimum value of f(x) on the interval .

Note that we did not need to include the endpoints in our investigation of . They are not needed in our procedure and need special handling even for definition since the derivative at such points can only make sense from one direction.

Example 4: Of all rectangles of area 100, which has the smallest perimeter?

The first step toward solving this problem is to reduce it to a pure math problem. If denotes one of the sides of the rectangle, then the adjacent side must be 100/x (in order that the area be 100). So the function we want
to minimize is

f(x) = p erimeter of an x by rectangle = .

The perimeter function is given by the formula 2x+200/x, where the domain is x>0 (because it makes no sense to talk of a rectangle with negative side). We have reduced the problem to finding the minimum value of f(x) for x between 0 and + .

We next find and set it equal to zero: = - / . To solve for x, we clear denominators and then move the constant to the left, obtaining [Maple Math] , which gives us (actually, = +/- 10, but does not have practical meaning). Since is defined for all x>0, so there are no corners or cusps to check. Our next step is to examine the behavior of at the endpoints. The sketch of this function = 2 +200/ in the region of positive is shown below at the right. We find that as gets close to 0 the function becomes larger and larger; and the perimeter function also becomes large when x gets very large.

code for perimeter graph

[Maple Plot]

(More precisely, the line x = 0 is a vertical asymptote , and the line is a slanted asymptote that the function approaches as -> .) Consequently, the minimum value of occurs at , and the desired rectangle is the 10 times 10-rectangle, whose perimeter is 40. Any other rectangle of area 100 has perimeter greater than 40. In the same way it is not hard to show that for any fixed area A, t he rectangle of area A with smallest perimeter is the square of side .

Another way to see that gives a minimum is to notice that decreases for a while as we get away from 0 ; but if we go too far to the right is increasing. In other words if ; while if . So it is clear that the point where really is a minimum.

Example 5: You want to sell a certain number x of items in order to maximize your profit. Market research tells you that if you set the price at $1.50, you will be able to sell 5000 items; and for every 10 cents you lower the price below $1.50 you will be able to sell another 1000 items. Suppose that your fixed costs (``start-up costs'') total $2000, and the per item cost of production (``marginal cost'') is $0.50. Find the price to set per item and the number of items sold in order to maximize profit , and also
determine the maximum profit you can get.

To answer these questions, we first observe that: profit is equal to the number of items sold times the price per item, minus the total cost . If P denotes the price and x the number sold then this profit is . (The term in parentheses is fixed cost plus number of items times per item cost.)

To find P in terms of x, we first note that the information given allows us to write x in terms of P: +1000 (i.e., 5000 plus 1000 times the number of multiples of 10 cents that P is below 1.50). Solving for P and simplifying gives: . Thus, the profit function that we want to maximize is:

= = .

We are now ready to set the derivative equal to zero: = from which we obtain . Thus, we plan to sell 7500 items, which we accomplish by lowering the price to 1.25. It costs us ( ) = to produce these items, and we take in 7 = 9375, leaving us a profit of f(7500) = $9375- $3625= $3625. This is the largest profit we can get under the assumptions of this problem.

code for profit graph

[Maple Plot]

Further analysis: We did not take all the usual steps and we now explain why they do not affect the answer here. For a complete answer, they are necessary!

What is the interval of values for ? Clearly and cannot be too large. By looking at the function, clearly cannot be too large as we cannot be interested in a negative profit function. Thus our interval could be [ ) and neither endpoint will be of interest, since it leads to negative profit. Since we have only one other point (zero of ) to consider, it must be the point with maximum profit.

It is also possible to find a more precise interval for corresponding to non negative profit, but it is hardly worth the effort!

To be safe, we should check that the value of x which made the profit-function have derivative 0 is really a maximum. It is not hard to compute that f(x) is increasing (i.e., ) when , and f(x) is decreasing when , so that f(x) reaches a peak at x = 7500. In most of the examples that follow (and when you do the homework problems) it is clear enough from the practical set-up that the value we find really is the maximum point (if we are looking for a maximum) or the minimum point (if we are looking for a minimum). So it is not necessary to belabor the matter. However, we should be cautious, as shown by Example 9 below.

Example 6: Find the largest rectangle that fits inside the graph of the parabola [Maple Math] below the line . (The top side of the rectangle should be on
the horizontal line y = a --- see the following diagram.)

[Maple Plot]

code for rectangle in parabola graph

Here the problem is to find where along the parabola to put the lower-right corner of the rectangle. That corner will determine all the other corners. Thus, if we let ( [Maple Math] ) be the point on the parabola where the lower-right corner is put, then the upper-right corner will be (x,a), the lower-left corner will be ( ), and the upper-left corner will be ( ). We can now write the area of the rectangle in terms of x:

= [Maple Math] .

To maximize this area function, we set

0 = [Maple Math]

i.e., [Maple Math] . The resulting rectangle has dimensions by , and its area is .

Additional considerations : It seems reasonable that the biggest possible rectangle should be situated with sides parallel to the x and y axes and corners on the parabola and the line y = a as stipulated. However, this has not been proved . We need to argue this carefully. If the rectangle is situated in a crooked manner, then we should be able to straighten it and then make it slightly bigger! You should think and try to establish these points .

Example 7: If you fit the largest possible cone inside a sphere, what fraction of the volume of the sphere is occupied by the cone? (Here by `cone' we mean a right circular cone, i.e., a cone for which the base is perpendicular to the axis of symmetry, and for which the cross-section cut perpendicular to
the axis of symmetry at any point is a circle.)

Let R be the radius of the sphere, and let r and h be the base radius and height of the cone inside the sphere. What we want to maximize is the volume of the cone: . Here R is a fixed value, but r and h can vary. Namely, we could choose to be as large as possible --- equal to R --- by taking the height equal to R; or we could make the cone's height h larger at the expense of making a little less than . See the side view depicted in the figure where we have situated the picture in a convenient way relative to the x- and y-axes, namely, with the center of the sphere at the origin and the vertex of the cone at the far left on the x-axis.

code for cone diagram

Notice that the function we want to maximize --- (1/3) [Maple Math] --- depends on two variables. This happens in many max/min problems. However, the two variables are related in some way --- by a certain condition of the form C(r,h) = constant. So our next step is to find such a condition, 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 can be read off the above drawing: the upper corner of the triangle, whose coordinates are ( ),
must be on the circle of radius R. That is,

[Maple Math] .

We can solve for h in terms of r or for r in terms of h. Let's do the latter. Then we substitute the result for r in (1/3) [Maple Math] , obtaining (1/3) .

Notice a simple trick that we used. We do not actually solve for r, but
write and then plug in our result into our function. If you try to solve for , then you get two roots and you will have to simplify the expression again.

Recall that is just a constant --- the only variable now is h. Let us take
[Maple Math] = . What happened to the (1/3) ? Well, we dropped this constant factor for convenience. The value of that makes (1/3) maximum is the same as the value of h that makes f(h) maximum, so we don't need to keep that constant factor when we take the derivative and set it equal to zero.

We're now ready to set = [Maple Math] . This gives us . In other words, we should, in fact, choose h a little larger than R, at the expense of having somewhat less than R. The exact value of r is determined by the condition above: .
Solving for here (after setting ) gives: .

Finally, the fraction of the sphere occupied by this cone is:

[Maple Math] = = =

The cone takes up a little less than 30 % of the sphere.

Alternative Method: We can use the technique of implicit differentiation to avoid solving for h or r as follows.
We wish to find a maximum value for (dropping as before), subject to the condition ( . Let us think of everything as a function of h and take derivatives with respect to h. From the derivative of the relation, we get: / = 0. By calculation of the derivative of and setting it to zero, we get:

Example 8: You are making cylindrical containers to contain a given volume. Suppose that the top and bottom are made of a material that is N times as expensive (cost per unit area) as the material used for the lateral side of the cylinder. Find (in terms of N) the ratio of height to base radius of the cylinder that minimizes the cost of making the containers.

Let us first choose letters for various things: for the height, for the base radius, and for the volume of the cylinder, and for the cost per unit area of the lateral side of the cylinder. Using the area formulas for the top and bottom together and for the lateral side, we find that the total cost of the material is

Here c and N are constants that are given to us, but r and h are variables. To eliminate one of the two variables, we use the ( it condition) that the volume must be V, i.e., . We use this relation to eliminate (we could eliminate , but it's a little easier if we eliminate , which appears in only one place in the above formula for cost). Thus we solve /( ) and substitute in the cost function. As a result, we consider the function:

( / )+ = / + .

(What happened to the c and the 2 in the above formula? For convenience,
we can factor out the constant 2c and drop it from the expression to be minimized, as in the previous example.) We now set = / + , giving = the (positive) cube root of /( ). Next, we know that = /( ), and so the ratio / is equal to

i.e., we choose the height to be times the base radius for the most economical container.

Example 9: Suppose you want to reach a point that is located across the sand from a nearby road. Suppose that the road is straight, and is the distance from to the closest point on the road. Let be your speed on the road, and let , which is less than , be your speed on the sand. Right now you are at the point , which is a distance from . At what point should you turn off the road and head across the sand in order to minimize your travel time to ?

Let x be the distance short of where you turn off, i.e., the distance from to . We want to minimize the total travel time. When you're traveling at constant velocity, you have:

= / .

In this problem (see the diagram at the right), we travel the distance DB at speed , and then the distance BA at speed . Since = and, by the Pythagorean theorem,

BA = , we want to minimize the time-of-travel function

code for diagram

[Maple Plot]

We set and next simplify by moving 1/ to the left and cross-multiplying:

0 = [Maple Math] , so that .

We now square both sides and solve for , obtaining

[Maple Math] , (i.e.,) .

This is our solution.

Notice, by the way, that our answer turns out not to depend on ; that is, the point where you should turn off the road does not depend on how far back you started. But something's not completely right here. What if is very small, i.e., suppose / [Maple Math] . Then our solution is saying that we should go ( it back) along the road, and then head across the sand. This makes no sense. What have we done wrong in the case of small ? The problem is with the first term ( )/ in our function . If that gives us negative time. What we really should have is / . In the case when / [Maple Math] , the graph of our revised function

[Maple Math]

code for diagram

[Maple Plot]

is given at the right. The minimum of this function cannot be obtained by setting (in fact, the derivative of this function is never 0). Rather, it occurs at a point where there is no derivative --- namely, at the ``corner'' at x = a. In practical terms, this says that if / [Maple Math] , then what you should do is head out immediately across the sand in a straight line to .

SUMMARY --- STEPS IN A MAX/MIN STORY PROBLEM:

1. Decide what the variables are and what the constants are, draw a diagram if appropriate, understand clearly what it is that is to be maximized or minimized.

2. Write a formula for what you are maximizing or minimizing.

3. Express that formula in terms of only one variable, that is, in the form .

4. Set and solve.

5. Be sure that your answer makes sense, and that the max/min value is not really somewhere else.

Be mindful that:

a. The max or min may be at an end end point

b. The max or min may be at a point where f(x) exists but does not exist.

c. There may be more than one point at which the max or min occurs

d. There may by no point at which the max or min occurs.

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