Optimization Problems

Problem-Solving Strategy

1. Understand the problem: Read carefully, identify what to maximize/minimize
2. Draw a diagram: Visualize the situation
3. Assign variables: Label all quantities
4. Write the objective function: Express what you're optimizing
5. Write constraint equation(s): Express relationships between variables
6. Reduce to one variable: Use constraints to eliminate variables
7. Find critical points: Set derivative = 0
8. Verify it's a max/min: Use first or second derivative test
9. Answer the question: Include units

Common Formulas

Geometry

• Rectangle: $A = lw$, $P = 2l + 2w$
• Circle: $A = \pi r^2$, $C = 2\pi r$
• Box: $V = lwh$, $S = 2lw + 2lh + 2wh$
• Cylinder: $V = \pi r^2 h$, $S = 2\pi r^2 + 2\pi rh$
• Cone: $V = \frac{1}{3}\pi r^2 h$

Distance

• $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$

Examples

Example 1: Fencing Problem

A farmer has 400 ft of fencing to enclose a rectangular field next to a river (no fence needed along river). Find dimensions for maximum area.

Let $x$ = side perpendicular to river, $y$ = side parallel to river.

Constraint: $2x + y = 400$, so $y = 400 - 2x$

Objective: $A = xy = x(400 - 2x) = 400x - 2x^2$

$\frac{dA}{dx} = 400 - 4x = 0$

$x = 100 \text{ ft}$

$y = 400 - 200 = 200 \text{ ft}$

Check: $\frac{d^2A}{dx^2} = -4 < 0$, so this is a maximum.

Maximum area = $100 \times 200 = 20{,}000$ ft²

Example 2: Box with Maximum Volume

An open-top box is made from a 12×12 inch square by cutting squares from corners. Find cut size for maximum volume.

Let $x$ = side of cut squares

Dimensions: $(12-2x) \times (12-2x) \times x$

$V = x(12-2x)^2 = x(144 - 48x + 4x^2) = 4x^3 - 48x^2 + 144x$

$\frac{dV}{dx} = 12x^2 - 96x + 144 = 12(x^2 - 8x + 12) = 12(x-2)(x-6) = 0$

$x = 2$ or $x = 6$

Since $0 < x < 6$, we have $x = 2$ inches.

$V(2) = 2(8)^2 = 128$ in³

Maximum volume = 128 cubic inches

Example 3: Minimum Distance

Find the point on $y = x^2$ closest to $(0, 3)$.

Point on parabola: $(x, x^2)$

Distance squared: $D = x^2 + (x^2 - 3)^2$

(Minimizing $D^2$ also minimizes $D$)

$\frac{dD}{dx} = 2x + 2(x^2 - 3)(2x) = 2x + 4x^3 - 12x = 4x^3 - 10x = 2x(2x^2 - 5)$

$2x(2x^2 - 5) = 0$

$x = 0 \text{ or } x = \pm\sqrt{\frac{5}{2}}$

Compare values:

• $D(0) = 0 + 9 = 9$
• $D\left(\pm\sqrt{\frac{5}{2}}\right) = \frac{5}{2} + \left(\frac{5}{2} - 3\right)^2 = \frac{5}{2} + \frac{1}{4} = \frac{11}{4}$

Closest point: $\left(\pm\sqrt{\frac{5}{2}}, \frac{5}{2}\right) \approx (\pm 1.58, 2.5)$

Example 4: Minimum Cost

A cylindrical can must hold 1000 cm³. Material costs $0.01/cm² for sides and $0.02/cm² for top and bottom. Find dimensions for minimum cost.

$V = \pi r^2 h = 1000, \quad \text{so } h = \frac{1000}{\pi r^2}$

$\text{Cost} = 0.02(2\pi r^2) + 0.01(2\pi rh)$

$= 0.04\pi r^2 + 0.02\pi r \cdot \frac{1000}{\pi r^2}$

$= 0.04\pi r^2 + \frac{20}{r}$

$\frac{dC}{dr} = 0.08\pi r - \frac{20}{r^2} = 0$

$r^3 = \frac{20}{0.08\pi} = \frac{250}{\pi}$

$r = \sqrt[3]{\frac{250}{\pi}} \approx 4.30 \text{ cm}$

$h = \frac{1000}{\pi \cdot (250/\pi)^{2/3}} \approx 17.21 \text{ cm}$

Example 5: Minimum Travel Time

A lifeguard at point A must reach a drowning swimmer at point B. The guard runs at 8 m/s on sand and swims at 2 m/s. Find the optimal entry point.

Setup: Guard is 50m along shore and 40m inland from the point closest to swimmer, who is 60m offshore.

Let $x$ = distance along shore to entry point.

$\text{Time} = \frac{\sqrt{40^2 + x^2}}{8} + \frac{\sqrt{60^2 + (50-x)^2}}{2}$

$\frac{dT}{dx} = \frac{x}{8\sqrt{1600 + x^2}} - \frac{50-x}{2\sqrt{3600 + (50-x)^2}} = 0$

This gives: $\frac{\sin(\theta_1)}{8} = \frac{\sin(\theta_2)}{2}$ (Snell's Law!)

Solving numerically: $x \approx 10.6$ m