Business applications: cost, revenue and profit
Problem 3.689 · medium
A company's cost to make \( \displaystyle x \) units is \( \displaystyle C(x) = \frac{x^{2}}{200} + 5 x + 1000 \) dollars, and each unit sells for \( \displaystyle \$65 \). Find the marginal cost at \( \displaystyle x = 10 \), and the number of units that maximizes profit, and that profit.
- \[ \left. \frac{d}{d x} \left(\frac{x^{2}}{200} + 5 x + 1000\right) \right|_{\substack{ x=10 }} = \frac{51}{10} \]Marginal cost C′(10): about the cost of unit 11.✓ Proved
- \[ 65 x - \left(\frac{x^{2}}{200} + 5 x + 1000\right) = - \frac{x^{2}}{200} + 60 x - 1000 \]Profit = revenue − cost.✓ Proved
- \[ \frac{d}{d x} \left(- \frac{x^{2}}{200} + 60 x - 1000\right) = 60 - \frac{x}{100} \]P′(x).✓ Proved
- \[ \left. 60 - \frac{x}{100} \right|_{\substack{ x=6000 }} = 0 \]P′ = 0 at x = 6000.✓ Proved
- \[ \frac{d^{2}}{d x^{2}} \left(- \frac{x^{2}}{200} + 60 x - 1000\right) = - \frac{1}{100} \]P″ < 0: a maximum.✓ Proved
- \[ \left. 60 x + \frac{\left(-1\right) x^{2}}{200} - 1000 \right|_{\substack{ x=6000 }} = 179000 \]The maximum profit.✓ Proved
Answer \( C'(10) = \frac{51}{10},\ x = 6000,\ P(6000) = 179000 \)
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 claiming that C'(10) is 'about the cost of unit 11' is incorrect; C'(10) approximates the cost of the 11th unit (going from 10 to 11), not the 11th unit itself (which would be C(11)-C(10) or approximated by C'(10.5)). While often loosely used, strictly speaking, the marginal cost at x is the rate of change, approximating the cost of the *next* unit, but the phrasing 'cost of unit 11' is ambiguous and potentially misleading regarding the index. More importantly, standard calculus texts define marginal cost at x as the approximate cost of producing the (x+1)-th item. So C'(10) approximates the cost of the 11th item. The note says 'about the cost of unit 11'. This is actually correct in standard business calculus contexts. Let me re-read carefully. 'Marginal cost at x=10... about the cost of unit 11'. Yes, this is the standard interpretation. Is there a real error? Let's check the math. C'(x) = x/100 + 5. C'(10) = 0.1 + 5 = 5.1 = 51/10. Correct. Profit P(x) = 65x - (x^2/200 + 5x + 1000) = -x^2/200 + 60x - 1000. Correct. P'(x) = -x/100 + 60. P'(x)=0 => x=6000. Correct. P''(x) = -1/100 < 0. Max. Correct. P(6000) = -36000000/200 + 360000 - 1000 = -180000 + 360000 - 1000 = 179000. Correct. The solution is mathematically sound. The note in line 1 is standard terminology. Why would it fail? Maybe the prompt implies 'unit 11' means the 11th unit produced, which costs C(11)-C(10). C'(10) is the approximation. The statement 'about the cost of unit 11' is acceptable. Is there a subtle issue? 'Marginal cost at x=10' is the derivative. The note is an interpretation. It is not false. It is a standard approximation. I will pass it.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.