Up: Notes for Math 10
Notes on Max-Min Problems:
A procedure for handling max/min problems is as follows:
Example: Of all rectangles with perimeter 100, find the dimensions of the
one with the largest area.
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). The table we developed
last class (when the limits at the endpoints are infinite) can help here.
- 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:
- Verify that the function is continuous and the interval closed.
Note that A(x) is continuous since it is a polynomial,
and the interval [0,50] is closed.
- Compute the derivative.
A'(x) = 50 - 2x
- Find where the derivative is zero or undefined.
A'(x) = 50 - 2x = 0 if and only if x = 25.
The derivative is always defined in this case. So the only critical point
is x = 25, which is inside the interval.
- 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 10
Created: Oct 20 1997 ---
Last modified: Mar 12, 1999 2:48:40 PM