Business applications: cost, revenue and profit
Problem 3.683 · medium
A company's cost to make \( \displaystyle x \) units is \( \displaystyle C(x) = \frac{x^{2}}{200} + 2 x + 1000 \) dollars, and each unit sells for \( \displaystyle \$42 \). Find the marginal cost at \( \displaystyle x = 10 \), and the number of units that maximizes profit, and that profit.
- \[ \left. \frac{d}{d x} \left(\frac{x^{2}}{200} + 2 x + 1000\right) \right|_{\substack{ x=10 }} = \frac{21}{10} \]Marginal cost C′(10): about the cost of unit 11.✓ Proved
- \[ 42 x - \left(\frac{x^{2}}{200} + 2 x + 1000\right) = - \frac{x^{2}}{200} + 40 x - 1000 \]Profit = revenue − cost.✓ Proved
- \[ \frac{d}{d x} \left(- \frac{x^{2}}{200} + 40 x - 1000\right) = 40 - \frac{x}{100} \]P′(x).✓ Proved
- \[ \left. 40 - \frac{x}{100} \right|_{\substack{ x=4000 }} = 0 \]P′ = 0 at x = 4000.✓ Proved
- \[ \frac{d^{2}}{d x^{2}} \left(- \frac{x^{2}}{200} + 40 x - 1000\right) = - \frac{1}{100} \]P″ < 0: a maximum.✓ Proved
- \[ \left. 40 x + \frac{\left(-1\right) x^{2}}{200} - 1000 \right|_{\substack{ x=4000 }} = 79000 \]The maximum profit.✓ Proved
Answer \( C'(10) = \frac{21}{10},\ x = 4000,\ P(4000) = 79000 \)
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. The reviewers disagree about how one step is explained; every verdict is in the receipt. 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
Every verdict on record (6)
qwen3.6:27b-mlx: pass 2026-10-18gpt-oss:20b: pass 2026-10-18qwen3.6:27b-mlx: pass 2026-10-16gpt-oss:20b: pass 2026-10-16qwen3.6:27b-mlx: fail (misleading) 2026-10-14 — The comment in line 1 incorrectly defines marginal cost as 'about the cost of unit 11'. Marginal cost C'(x) approximates the cost of producing the (x+1)-th unit, so C'(10) approximates the cost of the 11th unit, but the phrasing 'about the cost of unit 11' is imprecise and potentially confusing; more critically, it implies C'(10) is the cost of the 11th unit specifically, whereas it is the instantaneous rate of change. However, the standard interpretation is that C'(n) approximates the cost of the (n+1)th unit. The bigger issue is that the prompt asks for 'the marginal cost', 'the number of units', and 'that profit'. The solution provides these. Let's look closer at the 'misleading' aspect. Is C'(10) the cost of unit 11? No, C(11)-C(10) is the cost of unit 11. C'(10) is an approximation. Saying it is 'about the cost of unit 11' is actually a standard heuristic explanation. Is there an error? Let's check the math. C'(x) = x/100 + 2. C'(10) = 0.1 + 2 = 2.1 = 21/10. Correct. Profit P(x) = 42x - C(x). P'(x) = 42 - (x/100 + 2) = 40 - x/100. Set to 0: x=4000. P''(x) = -1/100 < 0. Max. P(4000) = 42(4000) - (4000^2/200 + 2(4000) + 1000) = 168000 - (80000 + 8000 + 1000) = 168000 - 89000 = 79000. The math is correct. The comment 'about the cost of unit 11' is a common pedagogical simplification. Is it misleading? It might be considered 'style' or 'misleading' if it suggests equality. But usually, 'marginal cost at x=10 is the approximate cost of producing the 11th unit' is the standard definition. The text says 'about the cost of unit 11'. This is acceptable. Wait, looking at the JSON output requirements. If it's sound, pass. Is there any subtle error? The problem asks for 'the number of units that maximizes profit'. x=4000. 'that profit'. 79000. 'marginal cost at x=10'. 2.1. Everything seems correct. Why would it fail? Maybe the comment in line 1 is technically wrong because marginal cost is the rate of change, not the cost of a specific unit, even approximately? No, the approximation is standard. Let's look for a 'style' issue. The comment is slightly informal. But is it 'misleading'? A student might think C'(10) = C(11)-C(10). That is false. C'(10) ≈ C(11)-C(10). The word 'about' covers this. I will mark it as pass unless I find a real error. Re-reading line 1: 'Marginal cost C′(10): about the cost of unit 11.' This is a standard interpretation. I will pass.gpt-oss:20b: pass 2026-10-14
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 2026-10-18 with SymPy 1.14.0.