Optimization
Problem 3.81 · medium
A rectangular field along a straight river needs no fence on the river side. With 200 m of fencing, what is the largest area that can be enclosed?
- Let x be the two sides perpendicular to the river; the side parallel is 200 − 2x. Area A(x) = x(200 − 2x), 0 < x < 100.
- \[ \frac{d}{d x} x \left(200 - 2 x\right) = 200 - 4 x \]A'(x).✓ Proved
- \[ \left. 200 - 4 x \right|_{\substack{ x=50 }} = 0 \]A'(x) = 0 at x = 50.✓ Proved
- \[ \frac{d^{2}}{d x^{2}} x \left(200 - 2 x\right) = -4 \]A'' < 0: a maximum.✓ Proved
- \[ \left. x \left(200 - 2 x\right) \right|_{\substack{ x=50 }} = 5000 \]50 m by 100 m.✓ Proved
Answer \( 5000 \)
Lines: 4 proved, 1 not checked. The answer was also checked a second way, without looking at the solution. The explanation has not been reviewed yet.
The full receipt
| Line | Status | Checked by | Detail |
|---|---|---|---|
| 1 | Not checked | — | a sentence; read, not computed |
| 2 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| 3 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| 4 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| 5 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| answer, a second way | ✓ Checked independently | sympy 1.14.0 + mpmath 1.3.0 | the objective sampled at 200,001 points of its interval tops out at the same value |
Proved: SymPy reduced the difference between the two sides to zero. Checked independently: a separate
method, named above, confirmed it. Checked numerically: the two sides agree at every sampled point, which is evidence,
not proof. Reviewed: a model or a person read it; that is all a sentence can have. Solution by
generator:structured/optimization, checked 2026-09-26 with SymPy 1.14.0.