Optimization

Part 1 of 7 — Setting Up Optimization Problems

The Strategy

1. Identify the quantity to maximize or minimize (the objective function)
2. Write an equation for it in terms of your variables
3. Use a constraint to eliminate a variable (reduce to one variable)
4. Find critical points of the objective function
5. Verify it's actually a max or min (use endpoints or Second Derivative Test)

Worked Example: Fencing Problem

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

Let $x$ = width (perpendicular to river), $y$ = length (parallel to river).

Objective: Maximize $A = xy$

Constraint: $2x + y = 200$ → $y = 200 - 2x$

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

$A'(x) = 200 - 4x = 0$ → $x = 50$

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

$y = 200 - 100 = 100$. Maximum area = $5000$ m$^2$.

Setting Up Optimization 🎯

Key Takeaways — Part 1

1. Always define your variables clearly
2. Write the objective function (what to optimize)
3. Use the constraint to reduce to one variable
4. Verify using the Second Derivative Test or endpoint analysis

Optimization

Part 2 of 7 — Geometric Optimization

Box Problem (Classic AP Question)

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

Objective: Maximize $V = x(12 - 2x)(8 - 2x)$

Domain: $0 < x < 4$

Expand: $V = x(96 - 24x - 16x + 4x^2) = 4x^3 - 40x^2 + 96x$

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

Using the quadratic formula: $x = \frac{20 \pm \sqrt{400-288}}{6} = \frac{20 \pm \sqrt{112}}{6}$

$x \approx 1.57$ or $x \approx 5.10$

Since $x < 4$, use $x \approx 1.57$. $V \approx 1.57(8.86)(4.86) \approx 67.6$ cubic units.

Geometric Optimization 🎯

Key Takeaways — Part 2

1. For minimizing distance, it's easier to minimize $D^2$ (avoids square roots)
2. Check your domain carefully for geometric problems
3. The box-cutting problem is a classic — know the setup

Optimization

Part 3 of 7 — Cost & Revenue Optimization

Business Applications

• Revenue: $R(x) = x \cdot p(x)$ where $p(x)$ is the price-demand function
• Profit: $P(x) = R(x) - C(x)$ (revenue minus cost)
• Marginal cost: $C'(x)$ — the cost of producing one more unit
• Maximum profit occurs where $R'(x) = C'(x)$ (marginal revenue = marginal cost)

Worked Example

A company sells widgets: demand is $p = 100 - 2x$ (price per widget when $x$ widgets are sold). Cost: $C(x) = 200 + 5x$.

$R(x) = xp = x(100-2x) = 100x - 2x^2$

$P(x) = R(x) - C(x) = 100x - 2x^2 - 200 - 5x = -2x^2 + 95x - 200$

$P'(x) = -4x + 95 = 0$ → $x = 23.75$

Since $x$ must be a whole number, check $x = 23$ and $x = 24$:

• $P(23) = -2(529) + 95(23) - 200 = 927$
• $P(24) = -2(576) + 95(24) - 200 = 928$

Maximum profit = $928 at$x = 24$ widgets.

Applied Optimization 🎯

Key Takeaways — Part 3

1. Profit = Revenue - Cost
2. Max profit where marginal revenue = marginal cost
3. Average cost is minimized where $\bar{C}' = 0$

Optimization

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

Cylinder with Fixed Surface Area

Minimize the surface area of a cylinder with volume $V = 1000$ cm$^3$.

$V = \pi r^2 h = 1000$ → $h = \frac{1000}{\pi r^2}$

$S = 2\pi r^2 + 2\pi r h = 2\pi r^2 + \frac{2000}{r}$

$S'(r) = 4\pi r - \frac{2000}{r^2} = 0$

$r^3 = \frac{500}{\pi}$ → $r = \left(\frac{500}{\pi}\right)^{1/3} \approx 5.42$ cm

This gives $h = 2r$ — the optimal cylinder has height equal to its diameter!

3D Optimization 🎯

Key Takeaways \u2014 Part 4

1. 3D optimization follows the same strategy: objective + constraint
2. Express surface area or volume in one variable using the constraint
3. For optimal cylinders: $h = 2r$ (with top) or $h = r$ (without top)

Optimization

Part 5 of 7 — Distance & Angle Optimization

Minimizing Travel Distance

A lifeguard at point $A$ on the beach must reach a swimmer at point $B$ in the water. She runs on sand at 8 m/s and swims at 2 m/s. Where should she enter the water?

This uses Snell's Law: the optimal path has $\frac{\sin \theta_1}{v_1} = \frac{\sin \theta_2}{v_2}$.

Optimization with Trigonometry

When angles are involved, express the objective function using trig and differentiate.

Practice Problems 🎯

Key Takeaways \u2014 Part 5

1. Some optimization problems involve geometry with angles or paths
2. Set up the objective function carefully, then use standard calculus techniques

Optimization

Part 6 of 7 — AP-Style Workshop

Mixed optimization problems similar to AP free-response questions.

AP-Style Problems 🎯

Workshop Complete!

Optimization is all about translating word problems into calculus.

Optimization — Review

Part 7 of 7 — Comprehensive Assessment

Optimization Checklist

1. \u2705 Define variables and draw a picture
2. \u2705 Write the objective function
3. \u2705 Use constraint to reduce to one variable
4. \u2705 Find critical points
5. \u2705 Verify max/min (Second Derivative Test or domain check)

Final Assessment 🎯

Optimization — Complete! \u2705

You have mastered:

• \u2705 Setting up objective functions and constraints
• \u2705 Geometric, business, and distance optimization
• \u2705 Verifying solutions using calculus tests