Optimization

Optimization problems use derivatives to find the largest or smallest value of a real quantity, such as the greatest area, the least cost or the shortest distance. Write the quantity as a function of one variable, find its domain, find the critical points, and check which one gives the maximum or minimum.

Before this lesson: Absolute Extrema, Second Derivative Test and Literal Equations

The idea behind optimization

Every optimization problem has two parts. The objective function is the quantity you want to make as large or as small as possible. The constraint is an equation that limits your choices, like a fixed amount of fencing or a required volume.

You use the constraint to write the objective function in terms of just one variable. Then it is an absolute extrema problem: find the critical points and compare.

How to solve optimization problems

1. Read the problem and draw a picture. Label the quantities with variables.
2. Write the objective function: the formula for the quantity to maximize or minimize.
3. Write the constraint, solve it for one variable, and substitute it into the objective function.
4. Find the domain: the values that make sense in the problem.
5. Differentiate, set the derivative equal to 0, and solve.
6. Check that you have a max or min: compare the endpoints (closed interval) or use the second derivative test.
7. Answer the question that was asked, with units.

Optimization examples

Example 1: Fencing along a river

A farmer has 240 m of fencing to make a rectangular field along a straight river. No fence is needed along the river. What dimensions give the largest area?

2x + y = 240

y = 240 - 2x

A(x) = x(240 - 2x) = 240x - 2x^2

A'(x) = 240 - 4x = 0 \Rightarrow x = 60

A(0) = 0, \; A(60) = 7200, \; A(120) = 0

y = 240 - 2(60) = 120

Answer: 60 m by 120 m (with the 120-m side along the river), for an area of 7,200 m².

Example 2: An open box

An open box is made from a 12-in. by 12-in. square sheet by cutting equal squares of side x from the corners and folding up the sides. What value of x gives the largest volume?

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

V'(x) = (12 - 2x)^2 + x \cdot 2(12 - 2x)(-2)

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

x = 6 \text{ or } x = 2

V(0) = 0, \; V(2) = 2(8)^2 = 128, \; V(6) = 0

Answer: cut 2-in. squares. The box is 8 in. by 8 in. by 2 in., with volume 128 in³.

Example 3: The cheapest can

A can (a closed cylinder) must hold 500 cm³. What radius uses the least metal?

\pi r^2 h = 500 \Rightarrow h = \frac{500}{\pi r^2}

S = 2\pi r^2 + 2\pi rh

S(r) = 2\pi r^2 + \frac{1000}{r}

S'(r) = 4\pi r - \frac{1000}{r^2} = 0

r^3 = \frac{1000}{4\pi} = \frac{250}{\pi}

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

S''(r) = 4\pi + \frac{2000}{r^3} \gt 0

h = \frac{500}{\pi(4.30)^2} \approx 8.60

Answer: a radius of about 4.30 cm (height about 8.60 cm). The best can is as tall as it is wide: h = 2r.

Example 4: The closest point on a curve

Find the point on the curve y = \sqrt{x} closest to the point (3, 0).

D^2 = (x - 3)^2 + (\sqrt{x} - 0)^2

D^2 = x^2 - 6x + 9 + x = x^2 - 5x + 9

\frac{d}{dx}\left[D^2\right] = 2x - 5 = 0 \Rightarrow x = 2.5

y = \sqrt{2.5} \approx 1.581

D = \sqrt{6.25 - 12.5 + 9} = \sqrt{2.75} \approx 1.658

Answer: the closest point is (2.5, √2.5) ≈ (2.5, 1.581), about 1.658 units away.

Example 5: Setting a ticket price

A theater sells 400 tickets at $20 each. For every $1 increase in price, it sells 10 fewer tickets. What price gives the most revenue?

R(x) = (20 + x)(400 - 10x)

R(x) = 8000 + 200x - 10x^2

R'(x) = 200 - 20x = 0 \Rightarrow x = 10

R''(x) = -20 \lt 0

R(10) = (30)(300) = 9000

Answer: charge $30 per ticket, for a revenue of $9,000 (300 tickets).

Common mistakes

Related methods

Minimizing distance

Distance has a square root, which makes the derivative messy. Since D is smallest exactly when D² is smallest, minimize D² instead, as in Example 4.

Business optimization

Profit is P(x) = R(x) − C(x). At the maximum profit, P′(x) = 0, so R′(x) = C′(x): the marginal revenue equals the marginal cost.

Practice problems

1. Find two positive numbers whose sum is 20 and whose product is as large as possible.
2. Find the largest area of a rectangle with perimeter 40 cm.
3. Find the smallest perimeter of a rectangle with area 64 m².
4. A farmer has 600 ft of fencing to make a rectangular pen split into two equal parts by a fence parallel to one side. Find the largest total area.
5. An open box with a square base must hold 32 ft³. Find the dimensions that use the least material.
6. Find two positive numbers whose product is 36 and whose sum is as small as possible.
7. Find the point on the line y = 2x + 1 closest to the origin.
8. A rectangle has its base on the x-axis and its upper corners on y = 12 - x^2. Find its largest area.
9. A rectangular field of 1,800 m² is fenced. One side costs $10 per meter and the other three cost $5 per meter. Find the dimensions with the lowest cost.
10. A company’s revenue is R(x) = 50x - 0.1x^2 and its cost is C(x) = 10x + 1000 dollars for x items. Find the maximum profit.

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

P'(x) = 20 - 2x = 0 \Rightarrow x = 10

2x + 2y = 40 \Rightarrow y = 20 - x

A(x) = x(20 - x), \; A'(x) = 20 - 2x = 0

P(x) = 2x + \frac{128}{x}

P'(x) = 2 - \frac{128}{x^2} = 0 \Rightarrow x^2 = 64

P''(8) = \frac{256}{512} \gt 0

3x + 2y = 600 \Rightarrow y = 300 - 1.5x

A(x) = x(300 - 1.5x) = 300x - 1.5x^2

A'(x) = 300 - 3x = 0 \Rightarrow x = 100

y = 300 - 150 = 150, \; A = 15{,}000

x^2h = 32 \Rightarrow h = \frac{32}{x^2}

S(x) = x^2 + 4xh = x^2 + \frac{128}{x}

S'(x) = 2x - \frac{128}{x^2} = 0 \Rightarrow x^3 = 64

h = \frac{32}{16} = 2, \; S = 16 + 32 = 48

S(x) = x + \frac{36}{x}

S'(x) = 1 - \frac{36}{x^2} = 0 \Rightarrow x = 6

D^2 = x^2 + (2x + 1)^2 = 5x^2 + 4x + 1

10x + 4 = 0 \Rightarrow x = -0.4

y = 2(-0.4) + 1 = 0.2

A(x) = 2x(12 - x^2) = 24x - 2x^3

A'(x) = 24 - 6x^2 = 0 \Rightarrow x = 2

A(2) = 4(8) = 32

C(x) = 10x + 5x + 5(2y) = 15x + \frac{18000}{x}

C'(x) = 15 - \frac{18000}{x^2} = 0 \Rightarrow x^2 = 1200

x = 20\sqrt{3} \approx 34.64, \; y \approx 51.96

P(x) = 50x - 0.1x^2 - 10x - 1000 = 40x - 0.1x^2 - 1000

P'(x) = 40 - 0.2x = 0 \Rightarrow x = 200

P(200) = 8000 - 4000 - 1000 = 3000

Optimization FAQ

What are the steps for solving optimization problems?

Draw and label, write the objective function, use the constraint to get one variable, find the domain, set the derivative equal to 0, check for a max or min, and answer the question.

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

On a closed interval, compare f at the critical points and endpoints. Otherwise use the second derivative test, or the first derivative test.

What is the constraint in an optimization problem?

It is the condition that limits your choices, such as a fixed perimeter, volume or budget. You use it to remove one variable.

Why minimize the distance squared?

The square root makes the derivative harder, and D is smallest exactly where D² is smallest, so minimizing D² gives the same point.

Summary

• Identify the objective function and the constraint.
• Use the constraint to write the objective in one variable, and find its domain.
• Set the derivative equal to 0 and solve.
• Confirm a max or min with endpoints or the second derivative test.
• Answer the question asked, with units.

Previous lesson: Curve Sketching

Next lesson: Newton’s Method

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