Up: Notes for Math 13
Notes on Max-Min Problems:
A procedure for handling max/min problems is as follows:
Example: Of all rectangles with perimeter 100, which has the greatest
Note that in step 6 it may turn out that the interval is not closed; for
example, the situation may make sense for all values of x.
When this happens, there is no guarantee that an optimum value exists. In
such a case, you must make an argument explicitly that demonstrates that
an optimum must exist (then you can use the theorem about open intervals
to claim that it must be at a critical point). For example, you might use
the 2nd derivative test to show that the point you obtain is of the
correct type (max or min) and then show that the values of the function
near the end points of the interval (or near plus or minus infinity if you
have an infinite interval) are not optimal.
- Rewrite the problem with one piece of information per line.
- Draw several pictures illustrating different possibilities.
Some possibilities include a 1 ´ 49
rectangle, a 25 ´ 25 rectangle
(square), and a 30 ´ 20 rectangle.
- Draw a generic picture and introduce variables.
The width and height are varying, so we draw an x ´ y
rectangle and let x be the width, and y the height. Then
the area, A, is A = xy and the perimiter
is P = 2x + 2y = 100.
- Explicitly write down "Mazimize ________" or "Minimize ________" and
find a formula for the quantity to be optimized.
Maximize area: A = xy
- Use information you know to eliminate all but one variable.
We know 2x + 2y = 100, so x +
y = 50, so y = 50 - x, hence
A = x(50-x). Now A is a function
of x alone.
- Determine the interval where the formula makes sense.
We need x > 0 and y >
0. The latter means 50 - x > 0, or
x < 50. Thus x must be in the
- Optimize the function on this interval:
- Find where the derivative is zero.
A'(x) = 50 - 2x = 0 if and only if x = 25.
- Find where the derivative is not defined
The derivative is always defined, in this case.
- Compute the function's value at these points and at the end points of
A(25) = 25(50-25) = 25 x 25 = 625
A(0) = 0(50-0) = 0
A(50) = 50(50-50) = 0
- Select the optimum value.
The maximum of 625 occurs at x = 25.
- Compute the quantity requested in the problem.
In this case, we want the dimensions of the rectangle. We have the
x value, so we still need y. Since y = 50 -
x, we have y = 50 - 25 = 25. So the
rectangle of perimeter 100 with the greatest area is a
25 ´ 25 rectangle.
Up: Notes for Math 13
Created: Oct 20 1997 ---
Last modified: Oct 27, 1998 3:49:33 PM