Applying Optimization Methods - CALCULUS - MATHEMATICS IN HISTORY - Mathematics for the liberal arts

Mathematics for the liberal arts (2013)

Part I. MATHEMATICS IN HISTORY

Chapter 4. CALCULUS

4.9 Applying Optimization Methods

I have 200 m of fencing available to create a rectangular pen to hold some sheep. What is the largest area of grass that can be enclosed by my fence? We can use calculus to find the answer.

The key to solving an applied optimization problem is to find a way to model the situation with a function. We know we can use derivatives to optimize functions, that is, to find maxima and minima. To get started with the correct function, it helps to begin with the proper question.

What is it that we are trying to maximize or minimize?

In this case, we want to maximize the area of grass enclosed by the rectangular sheep pen. We need a function that represents the area, and then we can apply calculus.

Figure 4.15 A sheep pen.

images

It is usually a good idea to draw a picture and label it, like Figure 4.15. Taking this as our picture, the area of the pen is simply

images

We would almost be ready to apply calculus, except for one impediment. The functions we have worked with so far have only one variable, but our area is in terms of both x and y. So there is one more step to resolve before we can continue. We need to eliminate one of the variables.

Luckily, we have one more relevant piece of information. We plan to build the pen using 200 m of fence. Nothing requires that we use all of the available fence, but we want the pen to be as large as possible, so it makes sense to use it all. Hence, the perimeter of our rectangle should be 200 m, and we can write this relationship as an equation:

images

Solving for y, we get

images

We can use this to eliminate y. If we substitute for y in our area formula, we get

images

Now we can apply calculus to this function. Taking the derivative, we get

images

Remember, maxima and minima occur at critical points. There are no places where the derivative is undefined (that is, no places where we do something like divide by zero or take the square root of a negative), so we don’t have to worry about that. That means the only possible critical points happen when the derivative is zero. Solving:

images

When x = 50, the area is A = x(100 – x) = 50(100 – 50) = 502 = 2500 m2. To verify that this corresponds to the maximum area, compare with the area at values on either side of 50, say 49 and 51.

x

A(x)

49

2499 m2

50

2500 m2

51

2499 m2

As we can see, x = 50 m corresponds to a maximum. Since y = 100 – x = 100 – 50 = 50 m, the largest pen that can be made from 200 m of fence is a 50 m × 50 m square.

We always need to check the critical points. It is easy to become lazy, since it is sometimes a fair bit of work to get from the initial statement of a problem, to a function, to the derivative, to the critical points. But it would be a shame to accidently minimize the value of something we intended to maximize simply because we forgot to check!

Other ways to verify that x = 50 m is the location of the maximum pen size would be to realize that A = 100 – x2 is a parabola opening downward (so our critical point marks the vertex of the parabola) or to graph the function on a calculator or computer. The method that we use to check is not so important as remembering to do the checking.

Values at the Boundary

What if you had been challenged to use 200 m of fencing to build a rectangular pen of minimum size? The answer is easy. Use 0 m of the fence to build a 0 m × 0 m pen with a total area of 0 m2.

What if I insisted, as part of the puzzle, that you use the entire 200 m of fence? Derivatives and critical points can’t tell you the answer because we already found that the only critical point happens when x = 50 m, and it corresponds to the maximum area.

Looking at a graph, you might conclude that there is no minimum, since the graph of A(x) is a parabola opening downward. But this is silly, because most of the graph has negative function values, and in the real world the area of a rectangular pen can never be negative. In Figure 4.16 we’ve augmented the graph to emphasize the positive region.

Figure 4.16 The area of a pen cannot be negative.

images

In mathematical language, this problem is said to be “constrained.” It only makes sense on the closed interval [0,100], and f(x) has two minima on the interval. It has a minimum at each end, at the “boundary” of the interval.

The solutions x = 0 m and x = 100 m make some practical sense, by the way. When x = 0 m, we have y = 100 – x = 100 m. Our rectangle has degenerated into two parallel runs of fence slapped together with no space between then. When x = 100 m, then y = 0 m, and the two strings of fence run the other way. Either solution gives a minimum area for the pen of 0 m2 and uses all of the fence.

You should convince yourself that this does not contradict Observation 1 on p. 254, that a maximum or minimum cannot occur when the derivative of a function is negative. When we made that observation, we depended on there being points to the left where the function would be higher and points to the right where the function would be lower. Obviously, when we come to the end of the interval there are no more points, and a minimum (as we have observed) or a maximum can occur.

In a similar way, a positive derivative will not prevent a maximum or minimum from occurring at the end of an interval. So Observation 2 is not contradicted either.

Optimization principle: For a function defined on a closed interval, maxima and minima may (only) occur at critical points as well as at the endpoints of the interval. We have to check the function at each of these places.

images EXAMPLE 4.27

A student has 200 m of fence available to make a garden (Figure 4.17). She wants the shape of the garden to be like the free-throw lane on a basketball court, a rectangle capped by a semicircle. What dimensions make the largest garden (or the smallest)?

Figure 4.17 A fenced garden.

images

Solution: Deciding how to label a picture can sometimes be a real challenge. For this figure, it probably makes sense to label the radius of the semicircle, since both the area and perimeter formulas for circles are given in terms of r. This forces one side of the rectangle to be 2r, and it is left to label the other side, which we have marked with y in this figure.

What are we trying to maximize? The area of the garden is the area of the rectangle plus the area of the semicircle, A = (2r)y + imagesπr2. This formula has two variables, so we need to eliminate one before we proceed.

To eliminate a variable, we use the other piece of information that we know, which is that we are to use 200 m of fence. The 200 m perimeter of the garden comes from three sides of the rectangle together with the semicircle:

images

Solve for either r or y by getting it alone on one side:

images

If we substitute this into the area formula, we get

images

This is a formula that we can optimize. Before we search for critical points, let’s first determine if there is an interval that constrains the problem. Clearly, r can’t be negative since it represents a real-world distance.

But how large can r become? Since the amount of fence is fixed, as r gets larger it can only be that y becomes smaller. Now, y can’t be negative, so the largest r would correspond to y = 0. To have a perimeter of 200 m, when y = 0 we would need 200 = 2r + πr, and solving, images.

So our function is constrained to the interval images.

To find critical points, we differentiate:

images

This derivative is never undefined, so critical points can only come from the derivative being zero. Solving:

images

Our optimization principle tells us that to finish, we need to check the function at this critical point and at the ends of the interval.

r

A(r)

0

0

200/(4 + π)

2800.5

200/(2 + π)

2376.8

The smallest garden occurs when r = 0 and y = 100, and it has an area of 0 m2. The largest garden comes from making images, which (if you check) also makes y ≈ 28 m, and it has an area of approximately 2800.5 m2.

images

EXERCISES

4.41 For each function find the critical points and classify each as a maximum, minimum, or neither.

a) y(x) = 7x + 10

b) y(t) = t(t – 4)

c) y(t) = (t – 1)(t – 4)

d) f(x) = x3x

e) f(x) = images

4.42 For each function and interval, find the points where the function reaches its maximum and minimum.

a) y(x) = 7x + 10 on [0,1]

b) y(t) = t(t – 4) on [0, 4]

c) y(t) = (t – 1)(t – 4) on [0, 4]

d) f(x) = 1/x on [1, 2]

e) f(x) = images on [2,4]

4.43 A rectangular pen runs next to a stream, so one side does not require a fence. Find the dimensions that maximize and minimize the area of the pen assuming 200 m of fence is used.

4.44 A rectangular pen runs along an inside comer of an existing (large rectangular) fence, so two sides do not require a fence. Find the dimensions that maximize and minimize the area of the pen assuming 200 m of fence is used.

4.45 A garden in the shape of the free-throw lane on a basketball court is built with one side against an existing wall, so that side needs no fence as in Figure 4.18. What are the dimensions that maximize the area of the garden?

Figure 4.18 A fenced garden against a wall.

images

Figure 4.19 A ladder goes around a comer.

images

4.46 Imagine carrying a ladder down a hallway when you come to a right-angle comer, as in Figure 4.19. Assume that the ladder is arriving from a hallway that measures 2 m across and entering a hallway that measures 3 m across. If the ladder is too long, it may not make the turn.

a) Taking the ladder to be the hypotenuse of a right triangle measuring y + 3 along one leg and x + 2 along the other, give a formula for the length of the ladder in terms of x and y.

b) Use similar triangles to relate x and y.

c) Substitute for y in the answer you got from part (a) to write the length of the hypotenuse as a function of x alone.

d) We can find the length of the longest ladder that fits around the comer by minimizing the function in part (c). This makes sense, because the ladder has to fit around the tightest part of the comer. But minimizing this function is inconvenient, because the function has a square root in it. It is not too hard to minimize the square of the function, however. Find the value of x that minimizes the square of the function from part (c).

e) How long is the longest ladder that can make the turn?

4.47 A wire 90 cm long is divided into three straight pieces of wire. Two pieces are the same length, x, which leaves the remaining piece of length 90 – 2x. Use the wire to form an isosceles triangle.

a) What is the height of the isosceles triangle as a function of x, taking the “odd” side as the base?

b) What is the area of the isosceles triangle, as a function of x?

c) What values of x maximize or minimize the area of the triangle?

4.48 A string consisting of n 9-volt batteries is connected in series to a 100 ohm circuit. Assume the current supplied (in amps) depends on the number of batteries, n, according to the formula images

a) How much current does a single 9-volt battery supply?

b) Find the derivative of I(n).

c) Find n where I′(n) = 0.

d) How many batteries would we use to get the maximum amount of current, and how much current will they provide?

e) How many batteries would we use to get the minimum amount of current (you may assume we use fewer than 100 batteries)?