# Optimization Practice Problems with Answers and Solutions

> 10 calculus optimization practice problems with answers and full solutions: fences, boxes, cans, closest points, revenue and the fastest route.

Source: https://practicecalculus.com/practice-problems/optimization
Updated: 2026-11-06

$$ f'(x) = 0 \;\text{ or undefined} \;\Rightarrow\; \text{test critical points and endpoints} $$

## At a glance

- **Problems:** 10 word problems, with answers and full solutions
- **Level:** Calculus 1 and AP Calculus AB, Unit 5
- **You need:** Critical points, the second derivative test and basic area and volume formulas
- **Main idea:** Write the quantity to optimize as a function of one variable, set its derivative to 0, then confirm max or min
- **Practice with new numbers:** [Optimization lesson](/learn/calculus-1/applications-of-derivatives/optimization)

## Key takeaways

- Use the constraint to write the quantity you want in terms of one variable before you differentiate.
- Always state the domain. On a closed interval, compare the critical points with the endpoints.
- Confirm the answer is a max or a min with the second derivative test or a sign chart, then answer the question that was asked.

These **optimization practice problems** ask for the largest or smallest value of something real: an area, a volume, a distance, a cost or a time. Set each one up yourself before you check. The answers come right after the list, then the full solutions.

Every problem follows the same steps:

1. Draw a picture and name the variables.
2. Write the quantity to optimize (the objective).
3. Use the constraint to write it in terms of one variable, and find the domain.
4. Set the derivative to $0$ and solve.
5. Confirm it is a max or a min, then answer the question with units.

For more on the method, read the [optimization problems guide](/blog/optimization-problems).

## Problems

**Numbers**

1. Two positive numbers add up to $20$. What is the largest possible product?
2. Two positive numbers have a product of $16$. What is the smallest possible sum?

**Fences and boxes**

3. **River fence.** A farmer has $240$ m of fencing to make a rectangular pen along a straight river. No fence is needed along the river. What dimensions give the largest area?
4. **Open box.** An open-top box is made from a $12$ in by $12$ in sheet of cardboard by cutting equal squares from each corner and folding up the sides. What size squares give the largest volume?
5. **Box with a square base.** An open-top box with a square base must hold $32{,}000$ cm³. What dimensions use the least material?
6. **Soup can.** A closed cylindrical can must hold $500$ cm³. What radius and height use the least metal?

**Geometry and distance**

7. **Rectangle in a semicircle.** A rectangle sits inside a semicircle of radius $10$, with its base on the diameter. What is the largest possible area?
8. **Closest point.** Which point on the curve $y = \sqrt{x}$ is closest to the point $(3, 0)$?

**Business and motion**

9. **Ticket price.** At a price of $p$ dollars, a theater sells $600 - 20p$ tickets. What price gives the most revenue?
10. **Rowboat.** You are in a rowboat $3$ km from a straight shore. You want to reach a town on the shore $8$ km down the coast from the closest point. You row at $2$ km/h and walk at $4$ km/h. Where should you land to arrive as soon as possible?

## Answers

| Problem | Answer |
| --- | --- |
| 1 | $10$ and $10$, product $100$ |
| 2 | $4$ and $4$, sum $8$ |
| 3 | $60$ m by $120$ m (long side along the river), area $7200$ m² |
| 4 | $2$ in squares, volume $128$ in³ |
| 5 | Base $40$ cm by $40$ cm, height $20$ cm, $4800$ cm² of material |
| 6 | $r = \sqrt[3]{\frac{250}{\pi}} \approx 4.30$ cm, $h = 2r \approx 8.60$ cm |
| 7 | $100$ (width $10\sqrt{2}$, height $5\sqrt{2}$) |
| 8 | $\left(\frac{5}{2}, \sqrt{\frac{5}{2}}\right) \approx (2.5, 1.58)$, distance $\frac{\sqrt{11}}{2} \approx 1.66$ |
| 9 | $p = 15$ dollars, revenue $4500$ dollars |
| 10 | Land $\sqrt{3} \approx 1.73$ km down the coast, total time $\approx 3.30$ h |

## Step-by-step solutions

**1.** Let the numbers be $x$ and $20 - x$, with $0 < x < 20$. Maximize $P = x(20 - x) = 20x - x^2$:

$$P' = 20 - 2x = 0 \quad\Rightarrow\quad x = 10$$

$P'' = -2 < 0$, so this is a max. The numbers are $10$ and $10$, and the product is $100$.

**2.** Let the numbers be $x$ and $\frac{16}{x}$, with $x > 0$. Minimize $S = x + \frac{16}{x}$:

$$S' = 1 - \frac{16}{x^2} = 0 \quad\Rightarrow\quad x = 4$$

$S'' = \frac{32}{x^3} > 0$, so this is a min. The numbers are $4$ and $4$, and the sum is $8$.

**3.** Let $x$ be each side perpendicular to the river. The side along the river is $240 - 2x$, with $0 < x < 120$. Maximize

$$A = x(240 - 2x) = 240x - 2x^2, \qquad A' = 240 - 4x = 0 \quad\Rightarrow\quad x = 60$$

$A'' = -4 < 0$, so this is a max. The pen is $60$ m by $120$ m, with area $7200$ m².

**4.** Let $x$ be the side of each square. The base is $12 - 2x$ on each side and the height is $x$, with $0 < x < 6$:

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

Use the product rule and factor:

$$V' = (12 - 2x)^2 - 4x(12 - 2x) = (12 - 2x)(12 - 6x)$$

$V' = 0$ at $x = 6$ (an endpoint, where $V = 0$) and $x = 2$. Since $V$ is $0$ at both ends and positive in between, $x = 2$ gives the max: $V = 2\cdot 8^2 = 128$ in³.

**5.** Let the base be $x$ by $x$ and the height $h$. The constraint is $x^2h = 32000$, so $h = \frac{32000}{x^2}$. The material is the base plus four sides:

$$S = x^2 + 4xh = x^2 + \frac{128000}{x}, \qquad S' = 2x - \frac{128000}{x^2} = 0$$

So $x^3 = 64000$ and $x = 40$. Then $h = \frac{32000}{1600} = 20$. Since $S'' = 2 + \frac{256000}{x^3} > 0$, this is a min, and $S = 1600 + 3200 = 4800$ cm².

**6.** A closed can has two circles and a side: $S = 2\pi r^2 + 2\pi rh$. The constraint $\pi r^2h = 500$ gives $h = \frac{500}{\pi r^2}$:

$$S = 2\pi r^2 + \frac{1000}{r}, \qquad S' = 4\pi r - \frac{1000}{r^2} = 0 \quad\Rightarrow\quad r^3 = \frac{250}{\pi}$$

So $r = \sqrt[3]{250/\pi} \approx 4.30$ cm. $S'' = 4\pi + \frac{2000}{r^3} > 0$, so it is a min. Then $h = \frac{500}{\pi r^2}$. Since $\pi r^3 = 250$, this is $h = \frac{2\pi r^3}{\pi r^2} = 2r \approx 8.60$ cm. The best can is as tall as it is wide.

**7.** Put the center of the semicircle at the origin. A corner of the rectangle sits at $(x, y)$ on the curve $y = \sqrt{100 - x^2}$. The width is $2x$:

$$A = 2x\sqrt{100 - x^2}, \qquad 0 \le x \le 10$$

$$A' = 2\sqrt{100 - x^2} - \frac{2x^2}{\sqrt{100 - x^2}} = \frac{2(100 - 2x^2)}{\sqrt{100 - x^2}} = 0 \quad\Rightarrow\quad x^2 = 50$$

So $x = 5\sqrt{2}$ and $y = \sqrt{50} = 5\sqrt{2}$. The area is $2(5\sqrt{2})(5\sqrt{2}) = 100$. At the endpoints $A = 0$, so this is the max.

**8.** A point on the curve is $(x, \sqrt{x})$. Minimize the squared distance, which has its min at the same place and is easier to differentiate:

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

The second derivative is $2 > 0$, so this is a min. The point is $\left(\frac{5}{2}, \sqrt{\frac{5}{2}}\right)$ and the distance is $\sqrt{\frac{1}{4} + \frac{5}{2}} = \frac{\sqrt{11}}{2} \approx 1.66$.

**9.** Revenue is price times tickets sold, with $0 \le p \le 30$:

$$R = p(600 - 20p) = 600p - 20p^2, \qquad R' = 600 - 40p = 0 \quad\Rightarrow\quad p = 15$$

$R'' = -40 < 0$, so it is a max. They sell $300$ tickets for $R = 4500$ dollars.

**10.** Land $x$ km from the closest point, with $0 \le x \le 8$. The rowing distance is $\sqrt{9 + x^2}$ and the walking distance is $8 - x$. Time is distance over speed:

$$T = \frac{\sqrt{9 + x^2}}{2} + \frac{8 - x}{4}, \qquad T' = \frac{x}{2\sqrt{9 + x^2}} - \frac{1}{4} = 0$$

So $2x = \sqrt{9 + x^2}$, which gives $4x^2 = 9 + x^2$ and $x = \sqrt{3} \approx 1.73$. Compare with the endpoints:

| Landing point | Time |
| --- | --- |
| $x = 0$ | $1.5 + 2 = 3.5$ h |
| $x = \sqrt{3}$ | $2 + \frac{3\sqrt{3}}{4} \approx 3.30$ h |
| $x = 8$ | $\frac{\sqrt{73}}{2} \approx 4.27$ h |

Landing $\sqrt{3}$ km down the coast is fastest.

## Common mistakes on these problems

- **Differentiating with two variables.** In problem 5, use the volume to replace $h$ before you take the derivative.
- **Skipping the endpoints.** On a closed interval like problem 10, the best answer could be at an end. Always compare.
- **Answering the wrong question.** Problem 4 asks for the size of the squares, so the answer is $2$ in, not just the volume.
- **Forgetting that the river side is free.** In problem 3 there are only three sides of fence: $2x + y = 240$.

## Keep practicing

These 10 problems use fixed numbers. For unlimited practice with new numbers and instant feedback on each step, try the [optimization lesson](/learn/calculus-1/applications-of-derivatives/optimization) and the [absolute extrema lesson](/learn/calculus-1/applications-of-derivatives/absolute-extrema). To see why the critical point method works, read about [Fermat's theorem](/theorems/fermats-theorem) and the [Extreme Value Theorem](/theorems/extreme-value-theorem), or practice finding extrema with the [curve sketching practice problems](/practice-problems/curve-sketching).

## Frequently asked questions

### How do you solve optimization problems in calculus?

Draw a picture and name the variables. Write the quantity to maximize or minimize, then use the constraint to reduce it to one variable. Find the domain, set the derivative to zero, test the critical points and endpoints, and answer in words with units.

### What is the constraint in an optimization problem?

It is the fixed condition, such as "240 m of fence" or "volume 500 cm³". It gives an equation that lets you solve for one variable in terms of the other, so the objective becomes a function of a single variable.

### How do you know if you found a maximum or a minimum?

Use the second derivative test: $f''(c) < 0$ means a local max and $f''(c) > 0$ means a local min. If there is only one critical point on an interval, that local max or min is also the absolute one. On a closed interval, you can also just compare the values at the critical points and endpoints.