Business applications: cost, revenue and profit
Problem 3.759 · medium
A company's cost to make \( \displaystyle x \) units is \( \displaystyle C(x) = \frac{x^{2}}{200} + 2 x + 1500 \) dollars, and each unit sells for \( \displaystyle \$42 \). Find the marginal cost at \( \displaystyle x = 100 \), and the number of units that maximizes profit, and that profit.
- \[ \left. \frac{d}{d x} \left(\frac{x^{2}}{200} + 2 x + 1500\right) \right|_{\substack{ x=100 }} = 3 \]Marginal cost C′(100): about the cost of unit 101.✓ Proved
- \[ 42 x - \left(\frac{x^{2}}{200} + 2 x + 1500\right) = - \frac{x^{2}}{200} + 40 x - 1500 \]Profit = revenue − cost.✓ Proved
- \[ \frac{d}{d x} \left(- \frac{x^{2}}{200} + 40 x - 1500\right) = 40 - \frac{x}{100} \]P′(x).✓ Proved
- \[ \left. 40 - \frac{x}{100} \right|_{\substack{ x=4000 }} = 0 \]P′ = 0 at x = 4000.✓ Proved
- \[ \frac{d^{2}}{d x^{2}} \left(- \frac{x^{2}}{200} + 40 x - 1500\right) = - \frac{1}{100} \]P″ < 0: a maximum.✓ Proved
- \[ \left. 40 x + \frac{\left(-1\right) x^{2}}{200} - 1500 \right|_{\substack{ x=4000 }} = 78500 \]The maximum profit.✓ Proved
Answer \( C'(100) = 3,\ x = 4000,\ P(4000) = 78500 \)
✓ Nihil obstat Every line of this solution was proved by the computer algebra system SymPy. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. How we verify
The full receipt
| Line | Status | Checked by | Detail |
|---|---|---|---|
| 1 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| 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 |
| 6 | ✓ 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 | profit tried at every whole number of units from 0 to 20000; marginal cost by a central difference |
Reviewers
gpt-oss:20b: passqwen3.6:27b-mlx: pass — The solution correctly derives the profit function, identifies the critical point, verifies the maximum using the second derivative test, and calculates the correct values for marginal cost, optimal units, and maximum profit.
Every verdict on record (6)
qwen3.6:27b-mlx: pass 2026-11-01 — The solution correctly derives the profit function, identifies the critical point, verifies the maximum using the second derivative test, and calculates the correct values for marginal cost, optimal units, and maximum profit.gpt-oss:20b: pass 2026-11-01qwen3.6:27b-mlx: pass 2026-10-31 — The solution correctly derives the profit function, identifies the critical point, verifies the maximum using the second derivative test, and calculates the correct values for marginal cost, optimal units, and maximum profit.gpt-oss:20b: pass 2026-10-31qwen3.6:27b-mlx: pass 2026-10-31 — The solution correctly derives the profit function, identifies the critical point, verifies it is a maximum using the second derivative test, and calculates the correct values for marginal cost and maximum profit.gpt-oss:20b: pass 2026-10-31
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/marginal_analysis, checked 2026-11-01 with SymPy 1.14.0.