One of the most important applications of the derivative concept is to “optimization” problems, in which some quantity must be maximized or minimized. Examples of such problems abound in many areas of life. An airline must decide how many daily flights to schedule between two cities to maximize its profits. A doctor wants to find the minimum amount of a drug that will produce the desired response in one of her patients. A manufacturer needs to determine how often to replace certain equipment to minimize maintenance and replacement costs.
Our purpose in this section is to illustrate how calculus can be used to solve optimization problems. In each example, we will find or construct a function that provides a mathematical model for the problem. Then, by sketching the graph of this function, we will be able to determine the answer to the original optimization problem
by locating the highest or lowest point on the graph. The y-coordinate of this point will be the maximum value or minimum value of the function.
The example is very simple because the functions to be studied are given explicitly.
Maximizing an Area You want to plant a rectangular garden along one side of a house, with a picket fence on the other three sides of the garden. Find the dimensions of the largest garden that can be enclosed using 40 feet of fencing.
Solution:
Let us think about this problem before embarking on the solution. With 40 feet of fencing, you can enclose a rectangular garden along with your house in many different ways. Here are three illustrations: In Fig. 3(a), the enclosed area is 10 × 15 = 150 square feet; in Fig. 3(b), it is 16 × 12 = 192 square feet; and in Fig. 3(c), it is 32 × 4 = 128 square feet. Clearly, the enclosed area varies with your choice of dimensions. Which dimensions that total 40 feet yield the largest area? We next show how we can use calculus and optimization techniques to solve this problem.
Since we do not know the dimensions, the first step is to make a simple diagram that represents the general case and assign letters to the quantities that may vary. Let us denote the dimensions of the rectangular garden by w and x (Fig. 4). The phrase “largest garden” indicates that we must maximize the area, A, of the garden. In terms of the variables w and x,
A = wx.
The fencing on three sides must total 40 running feet; that is, 2x + w = 40. (2) We now solve equation (2) for w in terms of x:
w = 40 − 2x. (3) Substituting this expression for w into equation (1), we have A = (40 − 2x)x = 40x − 2x^2 .
We now have a formula for the area A that depends on just one variable. From the statement of the problem, the value of 2x can be at most 40, so the domain of the function consists of x in the interval (0, 20). Thus, the function that we wish to maximize is
A(x) = 40x − 2x^2 0 ≤ x ≤ 20.
Its graph is a parabola looking downward (Fig. 5). To find the maximum point, we compute
A'(x) = 0
40 − 4x = 0
x = 10.
Since this value is in the domain of our function, we conclude that absolute maximum of A(x) occurs at x = 10. Alternatively, since A(x) = −4 < 0, the concavity of the curve is always downward, and so the local maximum at x = 10 must be an absolute maximum. The maximum area is A(10) = 200 square feet, but this fact is not needed for the problem.] From equation (3) we find that, when x = 10, w = 40 − 2(10) = 20.
In conclusion, the rectangular garden with the largest area that can be enclosed using 40 feet of fencing has dimensions x = 10 feet and w = 20 feet.
