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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.

  1. Rewrite the problem with one piece of information per line.
    Perimeter 100
    Maximize Area

  2. Draw several pictures illustrating different possibilities.
    Some possibilities include a 1 49 rectangle, a 25 25 rectangle (square), and a 30 20 rectangle.

  3. 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.

  4. Explicitly write down "Mazimize ________" or "Minimize ________" and find a formula for the quantity to be optimized.
    Maximize area: A = xy

  5. 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.

  6. 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 interval [0,50].

  7. Optimize the function on this interval:

    1. 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.

    2. Compute the derivative.

      A'(x) = 50 - 2x

    3. 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.

    4. Compute the function's value at these points and at the end points of the interval.
      A(25) = 25(50-25) = 25 x 25 = 625
      A(0) = 0(50-0) = 0
      A(50) = 50(50-50) = 0

    5. Select the optimum value.
      The maximum of 625 occurs at x = 25.

    6. 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.

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.

Up: Notes for Math 10

Comments to: dpvc@union.edu
Created: Oct 20 1997 --- Last modified: Mar 12, 1999 2:48:40 PM