Making the best choice
Optimization problems ask for the best dimensions: the largest area, the least material, the lowest cost. Every problem has:
- an objective: what to maximize or minimize
- a constraint: what’s fixed or limited
Classic 1: Fixed perimeter, maximum area
You have 40 m of fencing for a rectangular pen. What dimensions give the maximum area?
- Constraint: 2l + 2w = 40 → l = 20 − w
- Objective: A = lw = w(20 − w) = −w² + 20w
- Vertex: w = −20/(2 × −1) = 10 → l = 10
- 10 m × 10 m, area 100 m²
Result: For a fixed perimeter, a square encloses the most area.
| Width | Length | Area |
|---|---|---|
| 5 | 15 | 75 |
| 8 | 12 | 96 |
| 10 | 10 | 100 |
| 12 | 8 | 96 |
Classic 2: Three-sided fence
Fence a rectangle against a barn wall with 60 m of fencing (no fence along the wall).
- 2w + l = 60 → l = 60 − 2w
- A = w(60 − 2w) = −2w² + 60w
- Vertex: w = 15, l = 30 → A = 450 m²
The side parallel to the wall is twice the other sides.
Classic 3: Fixed area, minimum perimeter
A rectangle must have an area of 64 m². Minimum fencing?
Result: a square, 8 m × 8 m, perimeter 32 m.
3D optimization
Fixed volume, minimum surface area:
- A rectangular box with a given volume uses the least material when it’s a cube.
- A cylinder (closed can) with a given volume uses the least surface area when its height equals its diameter (h = 2r).
Example: A can must hold 1,000 cm³. Optimal r: 2r = h and πr²h = 1,000 → 2πr³ = 1,000 → r ≈ 5.42 cm, h ≈ 10.84 cm.
Fixed surface area, maximum volume: again a cube.
Real-world twists
Real packaging isn’t always optimal, because of shelf fit, grip, branding, stacking and shipping. Optimization gives a mathematical ideal; design balances it with other needs.
Strategy
- Draw and label a diagram.
- Write the constraint and the objective.
- Use the constraint to write the objective in one variable.
- Find the max or min (vertex of a parabola, a table of values, graphing, or calculus).
- Answer with dimensions and units, and check that they make sense.
Worked examples
Example 1: 100 m of fencing, rectangular pen. Max area? 25 × 25 = 625 m².
Example 2: 80 m along a river (no fence on the river side). Max area? w = 20, l = 40 → 800 m².
Example 3: A box with a square base has volume 27 m³. Least surface area? A cube 3 × 3 × 3 → 54 m².
Common mistakes
- Not using the constraint to reduce to one variable.
- Forgetting which side has no fence.
- Giving the area without the dimensions (or vice versa).
Try it yourself
- 24 m of fencing, rectangle. Max area?
- 120 m of fencing, three sides (one side is a wall). Max area?
- A rectangle has area 49 cm². Minimum perimeter?
- Why are real cereal boxes not cubes?
Answers: 1) 36 m² (6 × 6) 2) 1,800 m² (30 × 60) 3) 28 cm (7 × 7) 4) e.g. they need to fit on shelves, be easy to hold and display branding.
Quick recap
- Optimization finds the max or min of an objective under a constraint.
- Fixed perimeter → square; fixed volume → cube (or h = 2r for a can).
- Use the vertex of a parabola to find the optimum.