Lecture 11 MaxMin Problems Maxima and Minima Problems

Lecture 11 Max-Min Problems

Maxima and Minima Problems of type: “find the largest, smallest, fastest, greatest, least, loudest, oldest, youngest, cheapest, most expensive, … etc. If problem can be phrased as in terms of finding the largest value of a function then one is looking for the highest and lowest points on a graph (if they exist)

There is not a one-step test to detect “highest” and lowest points on a graph. What we can do is detect where relative maxima and minima occur. Relative extrema occur at critical numbers (remember that end points are critical numbers)

Strategy for solving Max-Min: • Phrase problem as a function for which one is to find largest or smallest value • Find all critical numbers of the function (including end points) • Make a table of values of the function at the critical numbers • IF the function is defined on a closed interval then the largest (smallest) functional value in the table is the maximum (minimum) value of the function. • If not on a closed interval can still detect the local maxima and minima – may be able to determine if one of them is an absolute max or min by sketching the graph.

length = x length = f(x) = x 2 0 Max Value P(x) = 2 x + 2( ) Critical numbers = 0, 2, Function is on closed interval [0, 2]

Want x-value of highest point = Critical numbers = -1 (ep), 0 (ep), -2/3 Function is defined on closed interval so max value at critical number is max value of function. MAX

Length = f(x) – g(x) or g(x)-f(x) h(x) = f(x) – g(x) x

or (> 1. 5) or Critical numbers = -1. 5 (ep), Just because a derivative is 0 does not mean it is a critical number of the function under consideration – Here 1. 720359. . is not in the domain. MAX

Length y Length x Area = base + 4 sides = x*x + 4*x*y

Area = base + 4 sides = x*x + 4*x*y Looking for lowest point – will occur at x = critical number near 3. A ‘ = 0 if x = = This is an example where the domain is not a closed Interval but we can still determine that the min occurs at the one critical number

You can buy any amount of motor oil at $. 50 per quart. At $1. 10 you can sell 1000 quarts but for each penny increase in the selling price you will sell 25 fewer quarts. Your fixed costs are $100 regardless of how many quarts you sell. At what price should you sell oil in order to maximize your profit. What will be your maximum possible profit? Profit = Income - Costs = (number sold)*(selling price) – [ (number purchased)*(purchase price) + fixed costs] Let x = increase in price in pennies Profit = (1000 -25*x)(1. 10+. 01*x) - [ (1000 -25*x)*. 50 + 100]

Critical numbers = 0 (ep), 40 (ep), -5/. 5 = 10 Maximum profit We determine here that there is no need to consider x > 40 which gives us a closed interval to work on Optimum selling price = $1. 10 + (-. 01*10) = $1