Unit 4 - Objective 4 - Minimum and Maximum Applications

Any problem which says, find "the maximum", "the greatest", "the minimum", "the least", "the most economical", "the optimum values", or something similar is a maximum-minimum problem.

Similar to graphing, we are looking for relative minima and maxima. However, we are not always given an equation for our function so we need to write the expression for the function using the description given.

Since minima and maxima only happen at critical values, and we do not need to sketch the function, we find the critical values and pick the one which gives the minimum or maximum that we are trying to find. When the function has more than one critical value, we can usually make the correct choice by using the Second Derivative Test:

For a critical value x = a where f '(a) = 0 if:

1. f ''(a) > 0, then there is a relative minimum when x = a. This means the graph has a horizontal tangent line and is concave up when x = a.
2. f ''(a) < 0, then there is a relative maximum when x = a. This means the graph has a horizontal tangent line and is concave down when x = a.
3. f ''(a) = 0, then the test fails. Try something else (usually check increasing and decreasing).

Plan for solving minimum-maximum problems:

1. Write an equation for what needs to be made a maximum or minimum.
2. Get that equation in terms of one variable (if it is not that already).
3. Take the derivative of that equation (with respect to that variable).
4. Set the derivative equal to zero and find the Critical Values (those are all possible maximum and minimum points).
5. Those critical values maybe checked in the second derivative (this assures your getting the desired max or min).
6. Answer the question posed in the problem.

The efficiency of an engine is given by

e = 0.768s - 0.00004s³, s > 0

Find the Maximum efficiency.

Solution: Following the plan above, "maximum efficiency" is asked for. The equation for Efficiency is given. It is already in terms of ONE variable (s).

 Take derivative: = .768 - 0.00012s² Set equal to zero: 0 = 0.768 - 0.00012s² Solve for critical value: 0.00012s² = 0.768 s² = 0.768/0.00012 = 6400, s =±80 s = 80 (not s = -80, since given s > 0) When s = 80, the efficiency, e, is a maximum. That maximum e value is e = 0.768(80) + 0.00004(80)³ = 40.96

Find the maximum area of a rectangle whose perimeter is 20 feet.

Solution: The key words are: Maximum Area, following the plan above:

1. Write an equation for area of a rectangle. A = lw
2. A = lw is in terms of 2 variables (l and w). One letter will have to be replaced.
Perimeter = 20 feet was also given.

 2l + 2w = 20 Solve for l (could have solved for w). 2l = 20 - 2w l = (20 - 2w)/2 = 10 - w Substitute this in A = lw. A = (10 - w) w = 10w - w² Now A is in terms of one variable, w.

3. Take derivative: dA/dw = 10 - 2
4. Set equal to 0 and solve:
• 0 = 10 - 2w
• 2w = 10
• w = 5 feet (gives a maximum area)
5. The question is to "find maximum area".
• The maximum area happens when w = 5, l = 10 - w = 10 - 5 = 5 feet
• Therefore the maximum area is A = lw = (5) 5 = 25 square feet