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Lesson 4: Optimization Problems

Estimated time: 40-50 minutes

Learning Objectives

Optimization Strategy

Optimization Procedure

  1. Understand the problem: Draw a picture. Identify what to maximize or minimize.
  2. Assign variables: Write an equation for the quantity to optimize (the objective function).
  3. Use the constraint to eliminate a variable so the objective has one variable.
  4. Find the critical points and determine which gives the optimal value.
  5. Verify using the First or Second Derivative Test (or compare endpoint values).

Classic Optimization Problems

Example 1: Fencing Problem

A farmer has 200 m of fencing to enclose a rectangular field along a river (no fence needed along the river). What dimensions maximize the area?

Variables: Let x = side perpendicular to river, y = side along river. Constraint: 2x + y = 200, so y = 200 − 2x.

Objective: A = xy = x(200 − 2x) = 200x − 2x².

Optimize: A'(x) = 200 − 4x = 0 ⇒ x = 50. y = 100.

Verify: A''(x) = −4 < 0, confirming a maximum. Max area = 5000 m².

Example 2: Minimum Distance

Find the point on the curve y = √x closest to (3, 0).

Distance: D = √[(x − 3)² + x]. Minimize D² = (x − 3)² + x = x² − 5x + 9.

Optimize: d(D²)/dx = 2x − 5 = 0 ⇒ x = 5/2. y = √(5/2).

Closest point: (5/2, √(5/2)).

Example 3: Open-Top Box

An open-top box is made from a 12 × 12 sheet by cutting squares of side x from each corner. What x maximizes volume?

Dimensions: (12 − 2x) × (12 − 2x) × x, where 0 < x < 6.

Volume: V = x(12 − 2x)² = x(144 − 48x + 4x²) = 4x³ − 48x² + 144x.

Optimize: V' = 12x² − 96x + 144 = 12(x − 2)(x − 6) = 0. x = 2 (since x = 6 is not in domain).

Max volume = 2(8)² = 128 cubic units.

Key Takeaways

Check Your Understanding

1. Find two positive numbers whose sum is 100 and whose product is maximum.

Answer: x + y = 100, P = x(100 − x). P' = 100 − 2x = 0, x = 50. Both numbers are 50. Max product = 2500.

2. A cylindrical can must hold 500 cm³. Find the radius that minimizes the surface area.

Answer: V = πr²h = 500, so h = 500/(πr²). S = 2πr² + 2πrh = 2πr² + 1000/r. S' = 4πr − 1000/r² = 0. r³ = 250/π, r = (250/π)1/3 ≈ 4.30 cm.

3. Find the rectangle of maximum area inscribed in a semicircle of radius 5.

Answer: Width = 2x, height = √(25 − x²). A = 2x√(25 − x²). A' = 0 gives x = 5/√2. Width = 5√2, height = 5/√2. Max area = 25.

4. A box with a square base and open top must have volume 32,000 cm³. What base side minimizes material?

Answer: V = s²h = 32000, h = 32000/s². Surface = s² + 4sh = s² + 128000/s. S' = 2s − 128000/s² = 0, s³ = 64000, s = 40 cm, h = 20 cm.

Ready for More?

Practice

10 practice problems covering Module 5.

Practice Problems