Introduction to Optimization — AP® Calculus AB Notes

Businesses want the most profit, engineers the least material, farmers the most land from a roll of fence. Each of these is a search for an absolute maximum or minimum, which you already know how to find. What is new in optimization is the first step: turning a paragraph of words into a function you can differentiate.

The objective function

The objective function is the quantity you want to make as large or as small as possible, written as a function of one variable. Writing it takes three moves:

Name the quantity. What does the question want largest or smallest? Area, revenue, cost, distance.
Write a formula for it. Often it starts with two variables, like $A = xy$ .
Use the given facts to reduce it to one variable. A fixed amount of fence, a fixed product, a price rule: solve it for one variable and substitute.

Then find the domain: the values of the variable that make physical sense. Lengths, prices and amounts are not negative.

Left: a rectangular pen against a river, split by a middle fence; the three fences perpendicular to the river have length x, the far side has length y, and 3x + y = 600 feet. Right: the graph of A(x) = x(600 - 3x) for 0 ≤ x ≤ 200, zero at both ends, highest at (100, 30000). — **Figure 1.** The fence fixes $3x + y = 600$ . Substituting $y = 600 - 3x$ turns the area into a function of $x$ alone, on $[0, 200]$ . LevlPrep original diagram.

Worked example 1. A farmer has 600 feet of fence for a rectangular pen against a straight river, split in two by a middle fence parallel to the ends. No fence goes along the river. Find the largest area.

Quantity: area, $A = xy$ , with $x$ the length of each end (and of the divider) and $y$ the side opposite the river.

Reduce: the fence is $3x + y = 600$ , so $y = 600 - 3x$ and $A(x)$ $= x(600 - 3x)$ $= 600x - 3x^2$ .

Domain: $x \ge 0$ and $y = 600 - 3x \ge 0$ , so $0 \le x \le 200$ .

Critical point: $A'(x) = 600 - 6x = 0$ at $x = 100$ .

Justify: candidates $A(0) = 0$ , $A(100) = 30{,}000$ , $A(200) = 0$ . The largest is $A(100)$ .

Answer in context: ends of 100 feet and a long side of 300 feet give the largest area, 30,000 square feet.

Justifying the answer

Setting $A'(x) = 0$ finds a candidate, not an answer. You must show it is the absolute extremum you want. Two ways work:

Situation	Argument
Closed domain $[a, b]$	Candidates Test: compare $f$ at the endpoints and the critical points.
Any interval, one critical point	The only critical point is a relative maximum (or minimum), by the First or Second Derivative Test, so it is the absolute one.

Watch for the critical point being the wrong kind. For a shipping cost $C(x) = x + \frac{16}{x}$ on $[1, 8]$ , the critical point $x = 4$ gives the smallest cost, 8. If the question asks for the largest cost, the answer is at an endpoint: $C(1) = 17$ .

Revenue, a second model

Worked example 2. At a price of $p$ dollars, a theater sells $1200 - 40p$ tickets, $0 \le p \le 30$ . What price maximizes revenue?

Objective: revenue = price times tickets, $R(p)$ $= p(1200 - 40p)$ $= 1200p - 40p^2$ .

$R'(p) = 1200 - 80p = 0$ at $p = 15$ . Candidates: $R(0) = 0$ , $R(15) = 9000$ , $R(30) = 0$ .

A price of 15 dollars gives the most revenue, 9000 dollars.

The derivative also has a meaning on its own. $R'(10) = 400$ dollars per dollar: at a price of 10 dollars, raising the price by 1 dollar raises revenue by about 400 dollars. A positive $R'$ says the best price is higher still.

Numbers and distances

Pure-number problems use the same steps. If two positive numbers add to 20, their product is $P(x) = x(20 - x)$ ; it is largest at $x = 10$ , giving 100. For a distance, minimize the squared distance instead: it is smallest at the same point and has no square root to differentiate. For the point of $y = \sqrt{x}$ nearest $(3, 0)$ , minimize $D(x) = (x - 3)^2 + x$ : $D'(x) = 2x - 5 = 0$ at $x = \frac{5}{2}$ , and the distance is $\sqrt{D\left(\frac{5}{2}\right)} = \frac{\sqrt{11}}{2}$ . Remember to take the square root at the end.

Answer the question asked

The final line should answer in the problem's words and units: "100 feet by 300 feet, for 30,000 square feet," not " $x = 100$ ." If the question asks for dimensions, give all of them; if it asks for the maximum value, give the value, not where it happens.