Optimization: Maximum and Minimum Problems

Q: What are the steps for an optimization problem?

Draw and name variables, write the quantity to optimize, use the constraint to reduce it to one variable, find the domain, find critical points, justify the max or min, and answer the question asked.

Q: What is the constraint?

The condition that ties the variables together, such as a fixed perimeter or a fixed volume. It lets you write the target quantity in terms of a single variable.

Q: How do I justify that my answer is a maximum?

On a closed interval, compare the critical points with the endpoints. Otherwise, use the first or second derivative test, or the single-critical-point argument.

Q: What are the dimensions of the cheapest can of a given volume?

The height equals the diameter. Minimizing surface area 2πr² + 2πrh with πr²h fixed gives h = 2r.

Q: Why must I state the domain?

Physical limits — lengths cannot be negative, a cut cannot exceed the sheet — decide which critical points count and which endpoints to compare.

Quick answer

An optimization problem asks for the largest or smallest value of a quantity under some condition: the most area for a fixed length of fence, the least metal for a can of fixed volume. The method: name the quantity, use the constraint to write it as a function of one variable, find that function's domain and critical points, and justify that the answer is a maximum or minimum — by comparing endpoints or with a derivative test. Then answer the question that was asked, with units.

What you'll learn

Translate a word problem into a function to optimize
Use a constraint to reduce it to one variable, with a domain
Find and justify the absolute maximum or minimum

The method

Draw the situation and name the variables.
Write the quantity to optimize in terms of those variables.
Use the constraint to eliminate all but one variable.
State the domain of the resulting function.
Find the critical points.
Justify the maximum or minimum: compare with the endpoints on a closed interval, or use a derivative test.
Answer the question that was asked, with units.

Why the constraint does the work

Most problems start with two unknowns. A rectangular pen has a length and a width, and “make the area as big as possible” means nothing until something limits them. That something is the constraint: $240$ meters of fence, a $1000$ cm³ can.

The constraint is an equation between the variables, so it can be solved for one of them. Substituting leaves the target as a function of a single variable, and then the tools of the last few lessons apply directly. The constraint turns a question about two quantities into a question about one function.

Example 1 — Fence along a river

A farmer has $240$ m of fence for a rectangular field along a straight river. No fence is needed on the river side. What dimensions give the largest area?

Let $x$ be the length of each side perpendicular to the river and $y$ the side parallel to it. The target is $A = xy$ . The constraint is $2x + y = 240$ , so $y = 240 - 2x$ and

A(x) = x(240 - 2x) = 240x - 2x^2, \qquad 0 \le x \le 120

$A'(x) = 240 - 4x = 0$ at $x = 60$ . Compare: $A(0) = 0$ , $A(60) = 7200$ , $A(120) = 0$ . The best field is $60$ m by $120$ m, with area $7200$ m².

A(x) = x(240 − 2x)

The area as a function of x

Worked examples

Example 3 — The cheapest can

A closed cylindrical can must hold $1000$ cm³. What radius uses the least metal?

Surface area $S = 2\pi r^2 + 2\pi r h$ , with constraint $\pi r^2 h = 1000$ , so $h = \tfrac{1000}{\pi r^2}$ and

S(r) = 2\pi r^2 + \frac{2000}{r}, \qquad r > 0

$S'(r) = 4\pi r - \tfrac{2000}{r^2} = 0$ gives $r^3 = \tfrac{500}{\pi}$ , so $r \approx 5.42$ cm and $h \approx 10.84$ cm. Since $S''(r) = 4\pi + \tfrac{4000}{r^3} > 0$ , this single critical point is the absolute minimum. Notice $h = 2r$ : the best can is as tall as it is wide.

Common mistakes

Practice problems

Two positive numbers add to $20$ . What is the largest possible product?

Answer

$100$

Full solution

$P = x(20 - x)$ on $[0, 20]$ . $P' = 20 - 2x = 0$ at $x = 10$ ; the endpoints give $0$ . $P(10) = 100$ .
A rectangle has perimeter $40$ cm. What is its largest possible area?

Answer

$100$ cm², a $10$ by $10$ square

Full solution

$2x + 2y = 40$ gives $y = 20 - x$ , so $A = x(20 - x)$ . The maximum is at $x = 10$ .
A rectangular pen with area $200$ m² has fence on all four sides. What dimensions use the least fence?

Answer

About $14.14$ m by $14.14$ m, a square

Full solution

$P = 2x + \tfrac{400}{x}$ for $x > 0$ . $P' = 2 - \tfrac{400}{x^2} = 0$ at $x = \sqrt{200} \approx 14.14$ , and $P'' = \tfrac{800}{x^3} > 0$ , so it is the minimum. The pen is square.
An open box is made from a $10$ -inch by $16$ -inch sheet by cutting squares of side $x$ from the corners. Find the $x$ that maximizes the volume.

Answer

$x = 2$ inches

Full solution

$V = x(10 - 2x)(16 - 2x) = 4x^3 - 52x^2 + 160x$ on $[0, 5]$ . $V' = 12x^2 - 104x + 160 = 4(3x - 20)(x - 2)$ . Only $x = 2$ is in the domain; $V(2) = 144$ beats the endpoint values $0$ .
Find the point on the line $y = 2x + 1$ closest to the origin.

Answer

$\left(-\tfrac{2}{5}, \tfrac{1}{5}\right)$

Full solution

$D = x^2 + (2x + 1)^2 = 5x^2 + 4x + 1$ . $D' = 10x + 4 = 0$ at $x = -\tfrac{2}{5}$ , where $y = \tfrac{1}{5}$ .
A box with a square base and open top must hold $32$ cubic feet. What dimensions use the least material?

Answer

Base $4$ ft by $4$ ft, height $2$ ft

Full solution

$x^2 h = 32$ , and the area is $S = x^2 + 4xh = x^2 + \tfrac{128}{x}$ . $S' = 2x - \tfrac{128}{x^2} = 0$ at $x^3 = 64$ , $x = 4$ , so $h = 2$ .
A rectangle is inscribed under the parabola $y = 12 - x^2$ , with its base on the $x$ -axis. Find its largest area.

Answer

$32$

Full solution

With corners at $(\pm x, 12 - x^2)$ , the area is $A = 2x(12 - x^2) = 24x - 2x^3$ on $[0, \sqrt{12}]$ . $A' = 24 - 6x^2 = 0$ at $x = 2$ , and $A(2) = 32$ .
The cost of producing $x$ items is $C(x) = 500 + 10x + 0.02x^2$ . What production level minimizes the average cost per item, $\tfrac{C(x)}{x}$ ?

Answer

About $158$ items

Full solution

Average cost $\bar{C} = \tfrac{500}{x} + 10 + 0.02x$ . $\bar{C}' = -\tfrac{500}{x^2} + 0.02 = 0$ at $x^2 = 25\,000$ , $x \approx 158.1$ . $\bar{C}'' > 0$ , so it is a minimum.
A window is a rectangle topped by a semicircle, with perimeter $10$ m. Which radius lets in the most light (largest area)?

Answer

$r = \tfrac{10}{4 + \pi} \approx 1.40$ m

Full solution

With radius $r$ , width $2r$ and rectangle height $h$ : $2r + 2h + \pi r = 10$ , so $h = 5 - r - \tfrac{\pi r}{2}$ .

Area $A = 2rh + \tfrac{\pi r^2}{2} = 10r - 2r^2 - \tfrac{\pi r^2}{2}$ . $A' = 10 - 4r - \pi r = 0$ gives $r = \tfrac{10}{4 + \pi}$ , and $A'' < 0$ .
For the fence problem, a student answers " $x = 60$ ". What is missing?

Hint

What did the question ask for?

Answer

The question asked for the dimensions: $60$ m by $120$ m, and a justification that this is a maximum.

Full solution

$x = 60$ is one side; the other is $y = 240 - 2(60) = 120$ .

A complete answer also shows it is the maximum: $A(60) = 7200$ beats $A(0) = A(120) = 0$ .

Frequently asked questions

What are the steps for an optimization problem?