Optimization Practice Problems with Answers and Solutions

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.

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

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'' = -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'' = \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'' = -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:

Use the product rule and factor:

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:

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}:

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:

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:

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'' = -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:

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 \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 and the absolute extrema lesson. To see why the critical point method works, read about Fermat's theorem and the Extreme Value Theorem, or practice finding extrema with the curve sketching practice problems.