Optimization

Optimization

Part 1 of 7 — Setting Up Optimization Problems

Topic Overview

The 5-Step Optimization Strategy

$\boxed{\text{1. Variables} \to \text{2. Objective} \to \text{3. Constraint} \to \text{4. Critical pts} \to \text{5. Verify}}$

Worked Example: Rancher Fencing

A rancher has 200 m of fencing to enclose a rectangle along a river (no fence on the river side). Find the maximum area.

Let $x =$ width, $y =$ length along river.

Constraint: $2x + y = 200 \Rightarrow y = 200 - 2x$

Objective: $A(x) = x(200-2x) = 200x - 2x^2$

$A'(x) = 200 - 4x = 0 \quad \Rightarrow \quad x = 50$

$A''(x) = -4 < 0$ — concave down — maximum

$y = 200-100 = 100$. $\boxed{A_{\max} = 50 \times 100 = 5000 \text{ m}^2}$

Key Fact: The optimal rectangle along a wall always has the side parallel to the wall equal to twice the perpendicular side.

Practice — Setting Up 🎯

Identify the setup. 🔍

Solve. ✍️

Key Takeaways — Part 1

• Define variables clearly and draw a diagram
• The objective is what you maximize/minimize
• The constraint reduces to one variable
• Use the Second Derivative Test or endpoint analysis to verify
• AP Tip: Always state the domain of the objective function

Optimization

Part 2 of 7 — Geometric Optimization

Classic Geometry Problems

Worked Example: Open-Top Box

An open-top box is made by cutting squares of side $x$ from the corners of a $12 \times 8$ sheet and folding up.

$V(x) = x(12-2x)(8-2x)$

Domain: $0 < x < 4$

$V = 4x^3 - 40x^2 + 96x$ $V'(x) = 12x^2 - 80x + 96 = 4(3x^2 - 20x + 24)$

$x = \frac{20 \pm \sqrt{400-288}}{6} = \frac{20 \pm \sqrt{112}}{6}$

$x \approx 1.57$ (only solution in domain). $\boxed{V_{\max} \approx 67.6 \text{ cubic units}}$

AP Tip: When asked to "set up but do not solve," write $V(x)$, state the domain, and show $V'(x) = 0$. You earn full credit without solving the quadratic.

Worked Example: Closest Point

Find the point on $y = \sqrt{x}$ closest to $(3, 0)$.

Minimize $D^2 = (x-3)^2 + (\sqrt{x})^2 = (x-3)^2 + x$

$\frac{d}{dx}D^2 = 2(x-3) + 1 = 2x - 5 = 0 \quad \Rightarrow \quad x = \frac{5}{2}$

Point: $\left(\frac{5}{2},\, \sqrt{\frac{5}{2}}\right)$

Key Fact: Always minimize $D^2$ instead of $D$ — it avoids square roots and gives the same critical points.

Practice — Geometric Optimization 🎯

Identify the correct approach. 🔍

Calculate. ✍️

Key Takeaways — Part 2

• Box-cutting problems: $V = x(L-2x)(W-2x)$, domain $0 < x < \min(L,W)/2$
• Minimize $D^2$ for closest-point problems
• Inscribed rectangle under a curve: $A = 2x \cdot f(x)$
• Always check domain endpoints for absolute max/min

Optimization

Part 3 of 7 — Applied & Business Optimization

Business Terminology

$\boxed{\text{Max profit where } R'(x) = C'(x)}$

Worked Example: Widget Company

A company sells widgets at price $p = 100 - 2x$ per widget. Cost: $C(x) = 200 + 5x$. Find maximum profit.

$R(x) = x(100-2x) = 100x - 2x^2$ $P(x) = R(x) - C(x) = 100x - 2x^2 - 200 - 5x = -2x^2 + 95x - 200$ $P'(x) = -4x + 95 = 0 \quad \Rightarrow \quad x = 23.75$

Since $x$ must be whole: $P(24) = -2(576) + 95(24) - 200 = -1152 + 2280 - 200 = 928$.

$\boxed{P_{\max} = 928 \text{ at } x = 24 \text{ widgets}}$, i.e. $928

AP Tip: On the AP exam, optimization word problems may use business language. Know the formulas for $R$, $C$, $P$, and marginal quantities.

Practice — Applied Optimization 🎯

Classify the quantity. 🔍

Solve. ✍️

Key Takeaways — Part 3

• Profit $= R - C$; max profit when $R'(x) = C'(x)$
• Average cost $\bar{C} = C(x)/x$; minimize by setting $\bar{C}' = 0$
• Revenue $R(x) = x \cdot p(x)$ where $p(x)$ is the demand function
• When units must be integers, check both nearest whole numbers

Optimization

Part 4 of 7 — 3D Optimization (Cylinders & Cones)

3D Shape Formulas

Worked Example: Minimize Surface Area

A closed cylinder has volume $V = 1000$ cm$^3$. Find the radius that minimizes surface area.

Constraint: $\pi r^2 h = 1000 \Rightarrow h = \frac{1000}{\pi r^2}$

Objective: $S = 2\pi r^2 + 2\pi rh = 2\pi r^2 + \frac{2000}{r}$

$S'(r) = 4\pi r - \frac{2000}{r^2} = 0$ $4\pi r^3 = 2000 \quad \Rightarrow \quad r = \left(\frac{500}{\pi}\right)^{1/3} \approx 5.42 \text{ cm}$

Then $h = \frac{1000}{\pi(5.42)^2} \approx 10.84 \approx 2r$.

$\boxed{\text{Optimal closed cylinder: } h = 2r}$

Key Fact: The optimal closed cylinder always has $h = 2r$ (height equals diameter). For an open-top cylinder, the optimal ratio is $h = r$.

Practice — 3D Optimization 🎯

Identify the setup. 🔍

Calculate. ✍️

Key Takeaways — Part 4

• 3D optimization: same 5-step process with volume/surface area formulas
• Closed cylinder: optimal when $h = 2r$
• Open-top cylinder: optimal when $h = r$
• Always express $S$ or $V$ in one variable using the constraint

Optimization

Part 5 of 7 — Distance & Mixed Problems

Distance Optimization

When minimizing the distance from a point $(a,b)$ to a curve $y = f(x)$:

$\boxed{D^2 = (x-a)^2 + (f(x)-b)^2}$

Key Fact: Always minimize $D^2$ — it shares the same critical points as $D$ and avoids square roots.

Worked Example: Wire-Cutting

A wire of length 20 is cut into two pieces. One is bent into a square, the other into a circle. What cut minimizes total area?

Let $x =$ circumference of circle, $20-x =$ perimeter of square.

Circle: $r = \frac{x}{2\pi}$, area $= \frac{x^2}{4\pi}$

Square: side $= \frac{20-x}{4}$, area $= \frac{(20-x)^2}{16}$

$A(x) = \frac{x^2}{4\pi} + \frac{(20-x)^2}{16}$

$A'(x) = \frac{x}{2\pi} - \frac{20-x}{8} = 0$

$\frac{4x}{8\pi} = \frac{\pi(20-x)}{8\pi} \quad \Rightarrow \quad 4x = 20\pi - \pi x$

$x = \frac{20\pi}{\pi + 4} \approx 8.80$

Special Optimization Patterns

Practice — Mixed Problems 🎯

Choose correctly. 🔍

Solve. ✍️

Key Takeaways — Part 5

• Minimize $D^2$ for closest-point problems (avoids square roots)
• Wire-cutting/splitting problems: express total in one variable
• Closed interval: always check endpoints AND critical points
• Discrete constraints: check both nearby integers

Optimization

Part 6 of 7 — AP-Style Workshop

AP FRQ Optimization Patterns

Full Worked AP Problem

A rectangle has one side on the $x$-axis, the upper two vertices on $y = 4 - x^2$.

(a) Express the area $A$ in terms of $x$.

(b) Find the value of $x$ that maximizes $A$.

(c) Justify that your answer gives a maximum.

Solution (a): Vertices at $(\pm x, 0)$ and $(\pm x, 4-x^2)$. Width $= 2x$, height $= 4-x^2$. $A(x) = 2x(4-x^2) = 8x - 2x^3, \quad 0 < x < 2$

Solution (b): $A'(x) = 8 - 6x^2 = 0 \Rightarrow x^2 = 4/3 \Rightarrow x = \frac{2}{\sqrt{3}} = \frac{2\sqrt{3}}{3}$

Solution (c): $A''(x) = -12x$. At $x = 2\sqrt{3}/3$: $A'' = -12 \cdot \frac{2\sqrt{3}}{3} < 0$.

Since $A'' < 0$ at the critical point, $A$ is concave down there, so this gives a maximum.

$A_{\max} = 2 \cdot \frac{2\sqrt{3}}{3}\left(4 - \frac{4}{3}\right) = \frac{4\sqrt{3}}{3} \cdot \frac{8}{3} = \frac{32\sqrt{3}}{9}$

AP Tip: On justification, you must state the Second Derivative Test result AND explicitly conclude "maximum" or "minimum." Just computing $f''$ is not enough.

AP-Style Practice 🎯

AP steps. 🔍

AP Problem. ✍️

Key Takeaways — Part 6

• AP FRQs require complete justification for max/min
• "Write but do not solve" earns credit for setup
• Second Derivative Test is the standard justification
• On closed intervals, compare all candidates (critical points + endpoints)

Optimization

Part 7 of 7 — Comprehensive Assessment

Optimization Strategy Reference

Common AP Mistakes

Quiz Set 1 — Core Skills 🎯

Quiz Set 2 — Advanced Problems 🎯

Final review. 🔍

Final Challenge. ✍️

🎉 Topic Complete!

You've mastered Optimization:

Key Fact: The 5-step optimization strategy (variables → objective → constraint → critical points → verify) works for every optimization problem. On the AP exam, always justify your answer using the Second Derivative Test or endpoint comparison.