The Fence Optimization Problem

You're given a fixed length of fence and asked to enclose the largest possible rectangular area. The intuition might suggest a long thin rectangle gives "more land per fence" — but the math says the opposite: the square is always optimal. This is a classic application of single-variable optimization with a constraint.

The Physics — Just an Optimization

Let $x$ be one side of the rectangle and $y$ be the other. We have two equations:

• Constraint (fixed perimeter): $2x + 2y = 100 \;\Longrightarrow\; y = 50 - x$
• Objective (maximize): $A = xy$

Substitute the constraint into the objective to get $A$ as a function of one variable. Then take the derivative, set it to zero, and solve.

Step-by-Step Solution

Given: Total perimeter $P = 100\;\text{m}$.

Find: The dimensions $(x, y)$ that maximize the enclosed area.

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Step 1 — Write the constraint and the objective.

$\text{constraint:} \;\; 2x + 2y = 100 \;\;\Longrightarrow\;\; y = 50 - x$

$\text{objective:} \;\; A = xy$

Step 2 — Substitute the constraint into the objective.

$A(x) = x(50 - x) = 50x - x^{2}$

Now $A$ is a function of a single variable $x$, valid for $0 \le x \le 50$.

Step 3 — Differentiate $A(x)$.

$A'(x) = 50 - 2x$

Step 4 — Set $A'(x) = 0$ and solve.

$50 - 2x = 0$

$2x = 50$

$x = 25\;\text{m}$

Step 5 — Find the corresponding $y$.

$y = 50 - x = 50 - 25 = 25\;\text{m}$

So the optimal rectangle has $x = y = 25\;\text{m}$ — it's a square.

Step 6 — Compute the maximum area.

$A_{\max} = (25)(25) = 625\;\text{m}^{2}$

Step 7 — Confirm it's a maximum (not a minimum) using the second derivative.

$A''(x) = -2$

Since $A''(x) = -2 < 0$ everywhere, the function $A(x)$ is concave down. The critical point at $x = 25$ is therefore a global maximum on $[0, 50]$. ✓

Step 8 — Compare with other shapes to feel the difference.

• A long thin rectangle $40 \times 10$: $A = 400\;\text{m}^{2}$ (35% smaller!)
• A medium rectangle $30 \times 20$: $A = 600\;\text{m}^{2}$
• A square $25 \times 25$: $A = 625\;\text{m}^{2}$ ← maximum
• A long thin rectangle $45 \times 5$: $A = 225\;\text{m}^{2}$

The further from a square you go, the worse you do — and the area falls off symmetrically on either side of $x = 25$ (since $A(x)$ is a downward parabola).

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Answer: The optimal rectangle is a square with side length $25\;\text{m}$, giving a maximum area of $\boxed{A_{\max} = 625\;\text{m}^{2}}$. Of all rectangles with a fixed perimeter, the square encloses the most area.

Generalization

This result holds for any total perimeter $P$: the optimum is always $x = y = P/4$, with $A_{\max} = P^{2}/16$. For $P = 100$, that's $625\;\text{m}^{2}$. For $P = 200$, it would be $2500\;\text{m}^{2}$, and so on. (And if you allow any shape, not just rectangles, the circle wins overall — but that requires multivariable calculus.)

Try It

• Slide the width widget — watch the rectangle reshape and the area number update.
• The graph below shows $A(x) = 50x - x^{2}$ — a downward parabola with peak at $x = 25$.
• The HUD lights up "★ MAXIMUM" when you're at $x = 25$.
• Notice the symmetric falloff: $A(20) = A(30) = 600\;\text{m}^{2}$, $A(15) = A(35) = 525\;\text{m}^{2}$, etc.