Geometric Optimization Math Example 1

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Example 1

medium
A farmer has P=60P = 60 m of fence to enclose a rectangular paddock against a straight wall (so only 33 sides need fencing). Find the dimensions that maximise the area.

Solution

  1. 1
    Step 1: Let width =x= x (the two sides perpendicular to the wall) and length =y= y (parallel to wall). Constraint: 2x+y=602x + y = 60, so y=60−2xy = 60 - 2x.
  2. 2
    Step 2: Area A=xy=x(60−2x)=60x−2x2A = xy = x(60 - 2x) = 60x - 2x^2.
  3. 3
    Step 3: Complete the square or differentiate: A=−2(x2−30x)=−2(x−15)2+450A = -2(x^2 - 30x) = -2(x-15)^2 + 450. Maximum at x=15x = 15 m.
  4. 4
    Step 4: y=60−2(15)=30y = 60 - 2(15) = 30 m. Maximum area =15×30=450= 15 \times 30 = 450 m2^2.

Answer

Width =15= 15 m, Length =30= 30 m, Maximum area =450= 450 m2^2.
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.

About Geometric Optimization

Finding the best geometric configuration — the shape that maximizes area, minimizes perimeter, uses the least material, or achieves some other optimal outcome — subject to given constraints.

Learn more about Geometric Optimization →

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