Optimization Problems

What You’ll Learn

In this lesson you’ll learn how to use derivatives to solve real-world optimization problems: finding the best, biggest, smallest, cheapest, or most efficient value of something.

The Concept

Optimization is the practical payoff of everything you’ve learned about derivatives. The process:

1. Read the problem and identify what you’re maximizing or minimizing
2. Draw a picture if possible
3. Write the quantity to optimize as a function of one variable (use constraints to eliminate extras)
4. Find the domain (what values make physical sense)
5. Take the derivative, set it to zero, find critical points
6. Test critical points (and endpoints if on a closed interval)
7. Answer the question in context

Worked Example

Example 1: Fence against a wall

You have 400 feet of fencing to enclose a rectangular pen against a straight wall. The wall forms one side, so you only need fence for three sides. What dimensions maximize the area?

Let x = width (perpendicular to wall), y = length (along wall).

Constraint: 2x + y = 400, so y = 400 - 2x.

Area as a function of x:

$$A(x) = x(400 - 2x) = 400x - 2x^2$$

Domain: x must be between 0 and 200 (if x = 200, there’s no length left).

Take the derivative and set it to zero:

$$A'(x) = 400 - 4x = 0 \quad \Rightarrow \quad x = 100$$

Then y = 400 - 2(100) = 200.

Maximum area is 100 times 200 = 20,000 square feet. The second derivative A” = -4 is negative, confirming this is a maximum.

Example 2: Open-top box

A box with a square base and no top must hold 500 cm³. What dimensions minimize the surface area?

Let x = side of the square base, h = height. Volume constraint: x²h = 500, so h = 500/x².

Surface area (base + four sides, no top):

$$S(x) = x^2 + 4xh = x^2 + 4x \cdot \frac{500}{x^2} = x^2 + \frac{2000}{x}$$

Take the derivative:

$$S'(x) = 2x - \frac{2000}{x^2}$$

Set to zero:

$$2x = \frac{2000}{x^2} \quad \Rightarrow \quad 2x^3 = 2000 \quad \Rightarrow \quad x^3 = 1000 \quad \Rightarrow \quad x = 10$$

Then h = 500/100 = 5 cm. The box is 10 cm by 10 cm by 5 cm tall.

Example 3: Maximum profit

A company’s profit function is P(x) = -0.01x² + 50x - 200, where x is units produced.

$$P'(x) = -0.02x + 50 = 0 \quad \Rightarrow \quad x = 2500$$

P”(x) = -0.02, which is negative, so x = 2500 is a maximum. Maximum profit is P(2500) = -0.01(6250000) + 125000 - 200 = 62,300.

Real-World Application

Optimization is everywhere:

• Manufacturing: minimize material cost for a container that holds a required volume
• Business: find the price point that maximizes revenue (price times quantity, where quantity depends on price)
• Engineering: design the lightest beam that can support a given load
• Logistics: find the route that minimizes fuel cost
• Game design: balance resource costs so the game economy feels fair

You’ve Got This

Optimization problems are word problems, and the hardest part is the setup. Once you have the function of one variable, the calculus is the same as finding critical points. Draw a picture, label everything, write the constraint, eliminate a variable, differentiate, solve. The pattern is always the same. Practice the setup and the rest follows.

Quiz

The first step in optimization is usually:

• A.Take the second derivative
• B.Identify what to maximize or minimize
• C.Graph the function
• D.Find all critical points

In the fence problem (400 ft, wall on one side), the optimal width is:

• A.200 ft
• B.50 ft
• C.100 ft
• D.133 ft

Why do you use constraints in optimization?

• A.To make the problem harder
• B.To eliminate variables so you have a function of one variable
• C.To find the domain only
• D.Constraints aren't needed

A negative second derivative at a critical point means:

• A.Local minimum
• B.Inflection point
• C.No extremum
• D.Local maximum

For the open-top box holding 500 cm³, the optimal base side is:

• A.5 cm
• B.10 cm
• C.15 cm
• D.20 cm