Learn Index · Calculus Grade 11–12

Optimization problems

Finding the maximum or minimum value in a real situation.

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Calculus finds the best

Optimization problems ask for the biggest or smallest value of something: maximum volume, minimum cost, shortest time. Derivatives find these exactly.

The process

Step Do
1 Draw a diagram; define variables.
2 Write the objective function (what to optimize).
3 Write the constraint and use it to reduce to one variable.
4 Find the feasible domain.
5 Differentiate; find critical numbers.
6 Test critical numbers and endpoints.
7 Answer the question in context, with units.

Example 1: maximum volume box

A 30 cm × 30 cm sheet of cardboard has equal squares (side x) cut from each corner, and the sides are folded up to make an open box. Find x for the maximum volume.

  • V(x) = x(30 − 2x)² = 4x³ − 120x² + 900x
  • Domain: 0 < x < 15
  • V′(x) = 12x² − 240x + 900 = 12(x² − 20x + 75) = 12(x − 5)(x − 15)
  • x = 5 (x = 15 is an endpoint where V = 0)
  • V(5) = 5 × 20² = 2,000 cm³

Example 2: minimum material for a can

A closed cylinder must hold 500 cm³. Minimize the surface area.

  • Constraint: πr²h = 500 → h = 500/(πr²)
  • A = 2πr² + 2πrh = 2πr² + 1,000/r
  • A′ = 4πr − 1,000/r² = 0 → r³ = 250/π → r ≈ 4.30 cm
  • h = 500/(π × 4.30²) ≈ 8.60 cm → h = 2r

Example 3: maximum revenue

A theatre sells 400 tickets at $20; each $1 increase loses 10 sales. Price p = 20 + x.

  • R(x) = (20 + x)(400 − 10x) = −10x² + 200x + 8,000
  • R′(x) = −20x + 200 = 0 → x = 10
  • Price $30; revenue $9,000

Example 4: shortest distance

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

  • Minimize D² = (x − 3)² + x = x² − 5x + 9 (squaring avoids roots)
  • d/dx = 2x − 5 = 0 → x = 2.5
  • Point (2.5, √2.5) ≈ (2.5, 1.58)

Checking it’s a max or min

Use the second derivative test, or compare values at critical numbers and endpoints.

Worked examples

Example 1: Two numbers add to 20. Maximize their product. P = x(20 − x) → x = 10 → 100.

Example 2: A rectangle with perimeter 60. Max area? 15 × 15 → 225.

Example 3: A farmer fences three sides of a field along a river with 400 m. Max area? A = x(400 − 2x) → x = 100 → 20,000 m².

Common mistakes

  • Not reducing to one variable.
  • Forgetting endpoints.
  • Answering with x when the question asks for the area, cost, etc.

Try it yourself

  1. Two positive numbers have a product of 64. Minimize their sum.
  2. An open box from a 12 × 12 sheet: best corner square size?
  3. Revenue R(p) = 120p − 2p². Price for maximum revenue?
  4. A rectangle is inscribed under y = 12 − x² (above the x-axis, symmetric). Max area?

Answers: 1) 8 + 8 = 16 2) 2 (volume 128) 3) p = 30 4) x = 2 → width 4, height 8, area 32

Quick recap

  • Optimization: write an objective function and a constraint.
  • Differentiate, find critical numbers, and check endpoints.
  • Answer in context.