🎯⭐ INTERACTIVE LESSON

Optimization

Learn step-by-step with interactive practice!

← Back to Standard Lesson

Optimization - Complete Interactive Lesson

Part 1: Setting Up Optimization Problems

Optimization

Part 1 of 7 — Setting Up Optimization Problems

The Strategy

Identify the quantity to maximize or minimize (the objective function)
Write an equation for it in terms of your variables
Use a constraint to eliminate a variable (reduce to one variable)
Find critical points of the objective function
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

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

Part 2: Geometric Optimization

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

Part 3: Cost & Revenue Optimization

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: $P ($ (revenue minus cost)

Part 4: 3D Optimization (Cylinders & Cones)

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 $^3$ .

$V = \pi r^2 h = 1000$ →

Part 5: Distance & Angle Optimization

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 $\frac{\sin \theta_1}{v_1} = \frac{\sin \theta_2}{v_2}$ .

Part 6: AP-Style Workshop

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.

Part 7: Comprehensive Assessment

Optimization — Review

Part 7 of 7 — Comprehensive Assessment

Optimization Checklist

\u2705 Define variables and draw a picture
\u2705 Write the objective function
\u2705 Use constraint to reduce to one variable
\u2705 Find critical points
\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

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

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

P (x) = R (x) - C (x)

Marginal cost:

C'(x)

— the cost of producing one more unit

Maximum profit occurs where

R'(x) = C'(x)

(marginal revenue = marginal cost)

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

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

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

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

Optimization with Trigonometry

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

Practice Problems 🎯

Key Takeaways \u2014 Part 5

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