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:
Plan for solving minimum-maximum problems:
The efficiency of an engine is given by
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 =±80s = 80 (not s = -80, since given s > 0) |
When s = 80, the efficiency, e, is a maximum. That maximum e value is |
Find the maximum area of a rectangle whose perimeter is 20 feet.
Solution: The key words are: Maximum Area, following the plan above:
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. |