Optimization

01What an optimization problem is

Every optimization problem has the same anatomy: an objective — the quantity to maximize or minimize (area, volume, cost, distance, time) — and usually a constraint — a fixed resource linking the variables (so much fence, so much material, a point that must lie on a curve). The calculus part is Module 08: find critical points, confirm max or min. The hard part is translation — converting a paragraph of English into a function of one variable. That translation is a repeatable skill, not a talent, and the procedure below makes it mechanical.

02The six-step procedure

1. Draw a picture and label every quantity with a symbol.
2. Write the objective — the formula for what's being optimized.
3. Write the constraint — the equation from the fixed resource.
4. Substitute the constraint into the objective to get a function of one variable.
5. Find the domain — what values are physically possible? (Lengths $\ge 0$, etc.)
6. Optimize: critical points, then verify max vs. min, then answer the question that was actually asked.

Step 5 is the one everyone skips and shouldn't: the domain decides whether you use the closed interval method (compare endpoints) or a derivative test, and endpoints sometimes win.

03Worked example: the fence

You have 400 ft of fence for a rectangular pen against a river — no fence needed on the river side. Maximize the area.

Picture: width $x$ (two sides), length $y$ (one side). Objective: $A = xy$. Constraint: $2x + y = 400$, so $y = 400 - 2x$. Substitute:

$$A(x) = x(400 - 2x) = 400x - 2x^2, \qquad 0 \le x \le 200$$

Optimize: $A'(x) = 400 - 4x = 0 \Rightarrow x = 100$. Then $A'' = -4 < 0$: a maximum. Dimensions $100 \times 200$ ft, area $20{,}000$ ft². (Endpoints give $A = 0$ — sanity confirmed.) Note the answer is not a square: the free river side breaks the symmetry, which is why you compute instead of guessing.

04Worked example: the open box

Cut equal squares of side $x$ from the corners of a 12 in × 12 in sheet and fold up the sides. Maximize the volume.

Folding gives a base of $(12 - 2x) \times (12 - 2x)$ and height $x$:

$$V(x) = x(12 - 2x)^2, \qquad 0 \le x \le 6$$

$V'(x) = (12-2x)^2 + x\cdot 2(12-2x)(-2) = (12 - 2x)(12 - 6x)$, zero at $x = 6$ (endpoint, $V = 0$) and $x = 2$. So $x = 2$ in gives the max: $V = 2 \cdot 8^2 = 128$ in³. The product-plus-chain differentiation and the factored form are typical — factoring $V'$ beats expanding it.

05Verifying max vs. min

A critical point is a suspect, not a conviction. Three ways to close the case: (1) Second derivative test — fastest when $f''$ is simple ($A'' = -4 < 0$ above). (2) First-derivative sign chart — always works. (3) Closed interval comparison — when the domain is closed, just evaluate at the critical points and endpoints; no test needed. And one honest shortcut: if the domain is an open interval, the function is positive inside, zero (or $\to 0$/$\to\infty$ appropriately) at the ends, and there's exactly one critical point — that point must be the extremum. State the reasoning; don't just assert it.

06Minimizing distance

Find the point on $y = x^2$ closest to $(0, 3)$. The distance from $(x, x^2)$ to $(0,3)$ is $\sqrt{x^2 + (x^2-3)^2}$ — but square roots are miserable to differentiate, so use the standard trick: minimize the squared distance, which has the same minimizer:

$$D(x) = x^2 + (x^2 - 3)^2 \qquad D'(x) = 2x + 2(x^2-3)(2x) = 2x(2x^2 - 5)$$

Critical points $x = 0$ and $x = \pm\sqrt{5/2}$. Compare: $D(0) = 9$, $D\left(\pm\sqrt{5/2}\right) = \tfrac{5}{2} + \tfrac14 = \tfrac{11}{4}$. Closest points: $\left(\pm\sqrt{5/2},\, 5/2\right)$. The symmetric pair of answers reflects the symmetric setup — a useful sanity check.

07Cost, revenue, and the economics pattern

Business problems dress the same math in dollars. A cylindrical can must hold $500$ cm³; material for the top and bottom costs twice the side material. The constraint $\pi r^2 h = 500$ eliminates $h$ from the cost objective $C = 2\cdot 2\pi r^2 k + 2\pi r h k$, leaving $C(r)$ to minimize. In pure revenue problems, if raising a price $p$ loses customers linearly, revenue $R = p \cdot (\text{quantity}(p))$ becomes a downward parabola — maximize as usual. The pattern never changes: constraint kills a variable, derivative finds the balance point where the competing pressures (bigger base vs. taller side, higher price vs. fewer buyers) exactly cancel.

08Pitfalls and pro habits

The four ways good students lose points: (1) Answering the wrong question — the problem asks for the maximum area and you report the optimal width, or vice versa; reread the final sentence. (2) Skipping the domain — and missing that an endpoint wins or that a negative critical point is meaningless. (3) No max/min justification — a bare "$f'(x) = 0$ at $x=2$" earns partial credit at best. (4) Differentiating before eliminating — the objective must be a one-variable function first; that's what the constraint is for. Build the habit of ending every solution with a sentence: "so the maximum area is 20,000 ft², achieved at 100 ft × 200 ft."