Solved Math Exams

What Are Optimization Problems?

Optimization problems ask you to find the best value of something — the maximum or minimum of a quantity subject to some constraint.

The key insight: If you can write a formula for what you want to optimize, calculus can find where it's maximized or minimized.

The General Strategy

Draw a picture (if applicable) and label variables
Write an equation for the quantity you want to optimize (call it $Q$)
Use constraints to eliminate variables until $Q$ depends on only one variable
Take the derivative and set it equal to zero
Solve for the critical points
Verify it's a max or min (using first or second derivative test, or domain analysis)
Answer the question they actually asked!

Example 1: Classic Fencing Problem

A farmer has 100 meters of fencing to enclose a rectangular field against a barn (so only 3 sides need fencing). What dimensions maximize the area?

Step 1: Set up variables

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

Step 2: Write constraint and objective

Constraint (fencing): $2x + y = 100$

Objective (area): $A = xy$

Step 3: Eliminate a variable

From constraint: $y = 100 - 2x$

Substitute: $A(x) = x(100 - 2x) = 100x - 2x^2$

Step 4: Find critical points

$$A'(x) = 100 - 4x = 0$$ $$x = 25$$

Step 5: Find the other dimension

$y = 100 - 2(25) = 50$

Step 6: Verify maximum

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

$$\boxed{x = 25 \text{ m}, \quad y = 50 \text{ m}}$$

Example 2: Closest Point on a Parabola

What are the $x$-coordinates of the points on the parabola $x = 4 - y^2$ closest to the origin?

Step 1: Set up the distance

Distance from $(x, y)$ to origin: $D = \sqrt{x^2 + y^2}$

Step 2: Minimize $D^2$ instead (easier!)

$D^2 = x^2 + y^2$

Since we're on the parabola: $x = 4 - y^2$

So: $D^2 = (4 - y^2)^2 + y^2$

Step 3: Let $f(y) = (4-y^2)^2 + y^2$ and find critical points

$$f(y) = 16 - 8y^2 + y^4 + y^2 = y^4 - 7y^2 + 16$$

$$f'(y) = 4y^3 - 14y = 2y(2y^2 - 7)$$

$f'(y) = 0$ when $y = 0$ or $y^2 = \frac{7}{2}$

Step 4: Find corresponding $x$-values

For $y = 0$: $x = 4 - 0 = 4$

For $y^2 = \frac{7}{2}$: $x = 4 - \frac{7}{2} = \frac{1}{2}$

Step 5: Compare distances

At $(4, 0)$: $D^2 = 16$

At $(\frac{1}{2}, \pm\sqrt{7/2})$: $D^2 = \frac{1}{4} + \frac{7}{2} = \frac{15}{4}$

Since $\frac{15}{4} < 16$, the closest points have $x = \frac{1}{2}$.

$$\boxed{x = \frac{1}{2}}$$

Example 3: The Ladder Around a Corner

A ladder must be carried horizontally around a corner where a 3-foot-wide hallway meets a 4-foot-wide hallway. What is the maximum length ladder that can make the turn?

Step 1: Set up with angle

Let $\theta$ = angle the ladder makes with the 4-foot hallway.

The ladder touches both walls. Using similar triangles:

Part in narrow hallway: $\frac{3}{\sin\theta}$
Part in wide hallway: $\frac{4}{\cos\theta}$

Step 2: Total length

$$L(\theta) = \frac{3}{\sin\theta} + \frac{4}{\cos\theta} = 3\csc\theta + 4\sec\theta$$

We want the minimum of this (longest ladder that fits = minimum of all possible "blocking" positions).

Step 3: Find critical points

$$L'(\theta) = -3\csc\theta\cot\theta + 4\sec\theta\tan\theta$$

Set equal to zero: $$4\sec\theta\tan\theta = 3\csc\theta\cot\theta$$

$$\frac{4\sin\theta}{\cos^2\theta} = \frac{3\cos\theta}{\sin^2\theta}$$

$$4\sin^3\theta = 3\cos^3\theta$$

$$\tan^3\theta = \frac{3}{4}$$

$$\tan\theta = \sqrt[3]{\frac{3}{4}}$$

Step 4: Calculate length

Using $\tan\theta = \sqrt[3]{3/4} \approx 0.909$, we get $\theta \approx 42.3°$

$$L \approx 3(1.49) + 4(1.35) \approx 9.87 \text{ feet}$$

$$\boxed{L \approx 9.87 \text{ feet}}$$

Example 4: Minimizing Travel Time

You're 2 km offshore and need to reach a point 6 km down the shore. You can row at 3 km/h and walk at 5 km/h. Where should you land to minimize total travel time?

Step 1: Set up variables

Let $x$ = distance along shore from the nearest point to where you land.

Step 2: Write time function

Rowing distance: $\sqrt{4 + x^2}$ (by Pythagorean theorem, since you're 2 km offshore)

Walking distance: $6 - x$

Total time: $T(x) = \frac{\sqrt{4 + x^2}}{3} + \frac{6-x}{5}$

Step 3: Find critical points

$$T'(x) = \frac{x}{3\sqrt{4+x^2}} - \frac{1}{5} = 0$$

$$\frac{x}{3\sqrt{4+x^2}} = \frac{1}{5}$$

$$5x = 3\sqrt{4+x^2}$$

$$25x^2 = 9(4+x^2)$$

$$25x^2 = 36 + 9x^2$$

$$16x^2 = 36$$

$$\boxed{\text{Land 1.5 km down the shore from the nearest point}}$$

Example 5: Maximum Product with Fixed Sum

Find two positive numbers whose sum is 20 and whose product is as large as possible.

Step 1: Set up

Let the numbers be $x$ and $y$ with $x + y = 20$.

Step 2: Write objective

$P = xy$

From constraint: $y = 20 - x$

$P(x) = x(20 - x) = 20x - x^2$

Step 3: Optimize

$P'(x) = 20 - 2x = 0$

$x = 10$, so $y = 10$

$$\boxed{x = y = 10}$$

Insight: For a fixed sum, the product is maximized when the numbers are equal!

Tips for Optimization Problems

Minimize $D^2$ instead of $D$ — avoids square roots, same critical points
Draw a picture — label everything, find relationships
Check your domain — physical constraints matter ($x > 0$, etc.)
Verify max vs min — use second derivative or endpoint analysis
Answer what they asked — they might want $x$, or $y$, or the max value itself

Common Mistakes and Misunderstandings

❌ Mistake: Forgetting to eliminate variables

Wrong: Trying to optimize $A = xy$ with two unknowns.

Why it's wrong: You can only take derivatives with respect to one variable.

Correct: Use the constraint to write everything in terms of one variable first.

❌ Mistake: Finding critical points but not verifying max/min

Wrong: "$x = 5$ is a critical point, so that's the maximum."

Why it's wrong: Critical points can be maxima, minima, or neither. Also, the max/min might be at an endpoint.

Correct: Use second derivative test, first derivative test, or compare endpoint values.

❌ Mistake: Answering the wrong question

Wrong: Finding $x = 10$ and writing that as the answer when they asked for the maximum area.

Why it's wrong: They might want the dimensions, or the maximum value, or something else.

Correct: Re-read the question. Plug your answer back in if they want the optimized value.

Optimization Problems for MATH 139

Exam Relevance for MATH 139

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