Problem 3.650
mediumA company's cost to make \( \displaystyle x \) units is \( \displaystyle C(x) = \frac{x^{2}}{200} + 4 x + 1500 \) dollars, and each unit sells for \( \displaystyle \$64 \). Find the marginal cost at \( \displaystyle x = 50 \), and the number of units that maximizes profit, and that profit.
- \[ \left. \frac{d}{d x} \left(\frac{x^{2}}{200} + 4 x + 1500\right) \right|_{\substack{ x=50 }} = \frac{9}{2} \]Marginal cost C′(50): about the cost of unit 51.✓ Proved
- \[ 64 x - \left(\frac{x^{2}}{200} + 4 x + 1500\right) = - \frac{x^{2}}{200} + 60 x - 1500 \]Profit = revenue − cost.✓ Proved
- \[ \frac{d}{d x} \left(- \frac{x^{2}}{200} + 60 x - 1500\right) = 60 - \frac{x}{100} \]P′(x).✓ Proved
- \[ \left. 60 - \frac{x}{100} \right|_{\substack{ x=6000 }} = 0 \]P′ = 0 at x = 6000.✓ Proved
- \[ \frac{d^{2}}{d x^{2}} \left(- \frac{x^{2}}{200} + 60 x - 1500\right) = - \frac{1}{100} \]P″ < 0: a maximum.✓ Proved
- \[ \left. 60 x + \frac{\left(-1\right) x^{2}}{200} - 1500 \right|_{\substack{ x=6000 }} = 178500 \]The maximum profit.✓ Proved
Answer \( C'(50) = \frac{9}{2},\ x = 6000,\ P(6000) = 178500 \)
✓ 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. The full receipt · How we verify