Geometric Optimization Examples in Math

Start with the recap, study the fully worked examples, then use the practice problems to check your understanding of Geometric Optimization.

This page combines explanation, solved examples, and follow-up practice so you can move from recognition to confident problem-solving in Math.

Concept Recap

Finding the best geometric configuration (maximum area, minimum distance, etc.).

What rectangle with fixed perimeter has the most area? A square!

Read the full concept explanation →

How to Use These Examples

  • Read the first worked example with the solution open so the structure is clear.
  • Try the practice problems before revealing each solution.
  • Use the related concepts and background knowledge badges if you feel stuck.

What to Focus On

Core idea: Optimal shapes tend to have high symmetry; the sphere maximises volume for any given surface area.

Common stuck point: The constraints in a problem determine what 'optimal' means—always identify and list all constraints first.

Sense of Study hint: Try testing a few specific shapes that meet the constraint and compare their areas or perimeters. The pattern often reveals the optimum.

Worked Examples

Example 1

medium
A farmer has P = 60 m of fence to enclose a rectangular paddock against a straight wall (so only 3 sides need fencing). Find the dimensions that maximise the area.

Solution

  1. 1
    Step 1: Let width = x (the two sides perpendicular to the wall) and length = y (parallel to wall). Constraint: 2x + y = 60, so y = 60 - 2x.
  2. 2
    Step 2: Area A = xy = x(60 - 2x) = 60x - 2x^2.
  3. 3
    Step 3: Complete the square or differentiate: A = -2(x^2 - 30x) = -2(x-15)^2 + 450. Maximum at x = 15 m.
  4. 4
    Step 4: y = 60 - 2(15) = 30 m. Maximum area = 15 \times 30 = 450 m^2.

Answer

Width = 15 m, Length = 30 m, Maximum area = 450 m^2.
Geometric optimisation finds the best (maximum or minimum) configuration under constraints. Using the wall replaces one length of fencing, so the optimal rectangle is not a square here — it is twice as long as wide.

Example 2

easy
A rectangle has perimeter P = 40 cm. Using the formula maximum area = P^2/16, compute the maximum area and the dimensions of the optimal rectangle.

Practice Problems

Try these problems on your own first, then open the solution to compare your method.

Example 1

easy
Two rectangles have the same perimeter of 24 cm: one is 8 \times 4 cm and one is 6 \times 6 cm. Which has greater area? Does this match the P^2/16 maximum?

Example 2

hard
A rectangular box with a square base and no lid must have volume V = 32 cm^3. Find the dimensions that minimise the total surface area (base + 4 sides).
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Related Concepts

AreaPerimeter

Background Knowledge

These ideas may be useful before you work through the harder examples.

areaperimeter