Max/Min Word Problems
One of the most widely used applications of differentiation is solving problems that require the maximization or minimization of a certain parameter. As with Newton's method and linear approximation, there is a standard procedure for approaching a word problem involving maximization or minimization. And again, a motivating problem will make the discussion clearer, so let's use the following:
A farmer wishes to build a rectangular pen for his animals out of 40 feet of fencing. What should be the length and width of pen so that its area is as great as possible?
Many students will intuitively know the answer to this question, but by following the steps below we can prove it using calculus!
- Visualize the problem and assign variables.
We'll let l = length of the pen
and let w = width of the pen
Finally, we'll let A = area of the pen
- Write equation to maximize or minimize.
We want to maximize the area.
The area of a rectangle is equal to the product of the lengths of its sides, so we have as our equation to maximize.
- List constraints.
There are constraints on the possible values of l and w.
Both must be positive for the pen to be a rectangle, and both must be less than 40 one couldn't have a rectangular pen. Thus, we get the constraints
Furthermore, the pen must have a perimeter of 40 feet. Thus, we get the restriction
- Rewrite the equation in two variables.
Right now, we have the equation to maximize. This invovles three variables—A, l, and w—and we don't know how to deal with such equations. Instead, we can use one of our restrictions above to rewrite one variable in terms of another:
Now, we can substitute this back into our original equation to maximize:
- Differentiate to find critical values.
We now have an equation in terms of just two variables, so we can differentiate it...
...and then set the derivative equal to zero to find the critical values:
- Check to see if critical value is a max/min.
We can check to see whether the critical value we have found is a maximum or a minimum with either the first derivative test or the second derivative test.
First derivative test:
Since the derivative is positive, zero, and then decreasing, the critical point w=10 is a maximum.
Second derivative test: Since , .
Since the second derivative is less than zero at the critical point (since it is less than zero everywhere), the critical point w=10 is a maximum.
- Answer the question
After you've gone through this procedure and done all of the calculus, it's important to remember to answer the question that was asked, including using whatever units may be necessary. In this case, we would answer the question as follows:
The farmer should build the pen with a width of 10 feet and a length of 10 feet.
Note the exact form of the answer will vary from question to question, even if the work done is exactly the same. For example, this question could have asked what the maximum area of the pen would be. To answer that question, we'd find the maximum point in exactly the same way, then plug into the equation for area to find A=10x10=100.
While not required, it's prudent to answer the question using a sentence to make sure you're giving all of the information asked for.
Following this procedure step-by-step makes solving max/min problems simple. The hardest part of these problems is probably coming up with the appropriate variables, restrictions, and equation to maximize or minimize. The best way to get better at this is to practice with some of the problems below.
This page last updated 8 August, 2008 5:51 PM