Max-Min Problems

MAX-MIN PROBLEMS

Max-min problems, or optimization problems, are word problems dealing with finding the maximum or minimum values of a function. For example:

1)  What are the dimensions of the box with the greatest volume that can be made from 20

square feet of cardboard?

2)  If you want to construct a cylindrical tin can that will hold 10 fluid ounces of liquid,

what should its dimensions be in order to use the least amount of tin?

The solution to such problems requires an understanding of the calculus techniques used to determine the extrema of a function. Potential candidates for maxima and minima are the critical values of a function – those values where the first derivative of the function is zero or undefined (recall also that the function must be defined at these values). We can classify these candidates using the first derivative test or the second derivative test, although in our word problems it is often clear from the nature of the problem whether or not a critical value is likely to be the appropriate extrema. (** If these ideas are unfamiliar to you, see the handout titled “Curve Sketching” before trying to go on with this handout.**) Since we are looking for the absolute maxima or minima of the function, we must also examine the behavior of the function at or near the endpoints of its domain. If an endpoint is in the domain of the function, it could be the absolute extremum we are seeking. If, on the other hand, the endpoint is not in the domain, the behavior of the function near the endpoint could rule out the existence of the absolute extremum we are looking for.

The following procedure should serve as a helpful guide for attacking max-min problems. The steps will be discussed and demonstrated using examples.

GENERAL STEPS TO USE IN SOLVING MAX-MIN PROBLEMS

(1) Read the problem carefully, determine what you are asked to find and record the objective of the problem in a simple statement. Then reread the problem.

(2) Draw a diagram to model the problem, and label relevant constants and variables.

(3) Determine your objective function. The objective function is the function to be maximized or minimized. If you are asked to maximize the area, then an expression representing the area is your objective function.

(4) Write down any constraint equations. A constraint equation is an equation representing any extra conditions that must be satisfied. There may not always be extra constraints in the problem.

(5) Rewrite the objective function in terms of only one variable. Often this is accomplished by

algebraically manipulating constraint equations and substituting into the objective function.

(6) Determine the domain of your objective function.

(7) Take the first derivative of the objective function and set it equal to zero (or determine where it is undefined) to find your critical values.

(8) Check the critical values and the endpoints to find the required extrema.

(9) Answer the original question completely.

To more clearly understand each of these steps, let us look at a max-min problem step-by-step.

EXAMPLE I:

A farmer wants to build a rectangular fence that will enclose 120 square feet of area for his dog. Two opposite sides are to be made of wood at a cost of $5 per foot. The other two sides are to be made of wire at a cost of $6 per foot. What are the dimensions of the fence that will cost the least to make?

SOLUTION:

(1) The first step in the successful solution of any max-min problem is to carefully read the statement of the problem (maybe a few times!). The purpose of this initial reading is to get an overview of the problem, not to solve it. You may start to see a geometric or algebraic model for the situation being described, but before you begin to write equations or facts, just think about the problem. Establish exactly what the problem is asking you to find, and think about how the solution will be influenced by the given information.

In the problem above you should ascertain that “cost” is the quantity to be minimized. You may find it helpful to actually note that in the first line of your written solution with a comment like “minimize the cost of the rectangular fence”. To blindly attempt to write down equations and relationships with no goal in mind often results on frustration and confusion.

Now that you have a goal in mind, think about the problem and why such a minimum should exist, and how it might be influenced by other factors in the problem. For example, if you want to minimize cost, why not just build a small fence out of cheaper material? Well, first of all, this fence must enclose 120 sq. ft. of area, so that will keep it from being too small. Also, we are required to have two sides of wood, and two of wire, so we must use at least some of the more expensive material. But it would be reasonable at this point to assume that the sides made of the cheaper material would probably be longer than the other two. Taking a little bit of time to think about this type of thing will make the problem more real to you, and help you see how to solve it. It will also make it easier for you to tell whether the answer you finally arrive at is reasonable.

(2) Next draw a picture of the situation. Since the goal is to find the dimensions of the fence that will minimize the cost, we can label the dimensions (variables) in our picture as L for length and W for width.

L

W W

L

(3)  Now we need our objective function. We already saw that we need to minimize cost, so an expression for the cost of the fence is our objective function. The cost of the fence will be the sum of the costs of the four sides. If we let the lengths be wooden (totally arbitrarily), then the cost of one length will be since wood is $5 per foot and L represents the number of feet on this side. Similarly the cost of one width of the fence will be since the wire is $6 per foot and W represents the number of feet on this side

(4) The only extra condition in this problem is that the fence must enclose 120 square feet. (This is important because without it the minimum cost would clearly be $0 for a fence with ). This gives us one constraint equation. Since the area of the rectangle is found by multiplying the length times the width we have

(5) We need to get the cost equation in terms of just one variable. It doesn’t matter which one. Using the constraint equation, we can solve for one variable in terms of the other, and then substitute that into the cost equation. Solving for W we get .

Substituting into our cost equation yields: (dollars)

(6) The next step is to determine the domain for the cost function that is dictated by the restrictions of the word problem. Invariably, the domain permitted by the conditions of the “real” problem is not as generous as the domain permitted by the equation C. If we look only at our equation, we see that L can be any real number except 0. However, since we know L is actually the length of the side of a rectangle, we know it makes no sense for L to be negative, and so our domain is actually L > 0.

(7) Now we are ready to take the first derivative of the cost equation and find the critical value(s).

(taking the derivative) /
(setting it equal to 0) /
(solving for L) /
(remember L > 0) /

Notice also, that is undefined at , but this is outside the domain of the COST function. Thus our only critical value occurs where .

(8) By using either the first derivative test, or the second derivative test we can classify this point.

Let’s use the second derivative test:

When , the second derivative is positive is positive, thus is a local minimum for C. Before we can conclude that is the absolute minimum we must consider the action of the function at the ends of the domain. Remember, if an absolute minimum exists, it will either be at a critical value or at an endpoint of the domain (if the endpoints are in the domain). If the endpoint is not part of the domain, we still need to test the behavior of the function close to the endpoint since it may indicate that there is no absolute minimum in our domain. In our case the domain is . Since the endpoints are not part of the domain (the interval is not closed) we consider what happens as we approach the endpoints. As , C clearly gets very large due to the term with the L in the denominator. As , C also gets very large due to the 10L term. So near the endpoints C is certainly larger than it is at , hence is our absolute minimum.

(9)  Finally we answer the question. The value of L that minimizes cost is 12. Recalling that we solve for the width, . The fence should be 10 feet by 12 feet, with the 10 foot sides made of wire and the 12 foot sides made of wood.

Before going on to another example, consider two slight variations on this problem.

A) What if you had been asked to find the dimensions of the fence that would be the most expensive to make? STEPS (1) – (7) WOULD HAVE BEEN EXACTLY THE SAME. The only difference would come when we are evaluating our critical values and considering the endpoints.

At the point where , the cost of the fence is . By our analysis above, C gets very large as or as .

For example, take , then

or take , then .

By choosing values of L larger than 1200 or smaller than 0.1, I can make a more and more expensive fence that will still enclose 1200 square feet. Thus there is no absolute maximum cost. I can make the fence as expensive as I like.

B) In all practicality, the fenced area must be at least wide enough for the farmer’s dog to walk in. If L is too large, then the width of the fenced area is so small, the dog couldn’t even squeeze in; if L gets too small, then the dog still can’t fit in the pen. So a more realistic version of A) would be “what are the dimensions of the fence that would be the most expensive to make, if the fenced area must be at least 3 feet on each side?”. STEPS (1) – (5) AND (7) WOULD AGAIN BE THE SAME. However, in step (6) our domain would be . (L can be no larger than 40 since W must also be at least 3.) Now when we think about step (8), the endpoints are a part of our domain. So we must check three places for absolute maximum, , , and .

At , and .

At , and .

At , and .

Thus the most expensive fence that can be made meeting the specifications in the problem is one that is 3 foot by 40 foot, with the 3 foot sides made of wire and the 40 foot sides made of wood.

These three problems represent the sort of problems that could arise when one enters the “real world” to apply the techniques of this and other courses. Notice that the solution to the problem is not just a number. It is the clear, complete, justified response to the question posed.

EXAMPLE II:

The starting point of a race is on an island 4 miles offshore in a straight river. The finish point is on the river bank, 6 miles downstream. Bill can swim at a rate of 6 mph and cycle at a rate of 12 mph. Assuming Bill can have his bicycle waiting anywhere on the bank, (and neglecting the influence of the current in the river), what route should he take to get from the starting point to the finishing point as quickly as possible?

SOLUTION:

We need to minimize the time it takes to get to the finish. Swimming directly to the finish would certainly be the shortest distance, but he swims more slowly than he cycles. / (1) Read (and reread) the problem and determine your goal.

Start
Bicycle
4 mi.
X Finish / (2) Draw and label a diagram.
Since we are trying to minimize time, we need an expression for the total time it takes.
= +
= + / (3) Determine the objective function.
His rate swimming is 6 mph.
His rate cycling is 12 mph. / (4) Write down the constraint equations.
Recall that the distance = (rate)(time), hence
.
So . Using the Pythagorean theorem and our diagram, we see that .
Now . From the diagram we see that . Substituting into the objective equation yields: / (5) Write the objective function in terms of only one variable.
Certainly Bill would not have his bicycle further upstream than the starting point, nor further downstream than the finish, as that would add unnecessary distance (and thus time) to his race. He may decide to swim directly across the river () or directly to the finish (), or anywhere in between. So our domain is: / (6) Determine the domain of T.

Since / (7) Find the critical values by setting the first derivative equal to 0 and solving for x.
Since T is a continuous function on the closed interval [0,6], it must have an absolute minimum value. Since T has a critical value at , the absolute minimum will either be there or at the endpoints of the domain, x = 0 or x = 6.
When , T 1.08 hours.
When x = 0, T 1.17 hours.
When x = 6, T 1.20 hours.
So the absolute minimum occurs when
(about 2.31 miles) / (8) Check the critical values and endpoints of the domain.
Bill can get to the finish the fastest by having his cycle waiting miles downstream from the starting point. His race will take about 1.08 hours. / (9) Answer the question completely.

You should now put yourself to work solving problems. You can only learn to solve max-min problems through experience and lots of practice.