Alternatively, perhaps the function is $ C(x) = 5000 + 120x - 0.5x^2 $, and we want to minimize total cost, not per unit. But the problem says "cost per unit."

February 22, 2026

$Alternatively, perhaps the function is $ C(x) = 5000 + 120x - 0.5x^2 $, and we want to minimize **total cost**, not per unit. But the problem says "cost per unit."$

But assuming the function is correct as stated, and the goal is to minimize $ P(x) = \frac{5000}{x} + 120 - 0.5x $, set:
P'(x) = -\frac{5000}{x^2} - 0.5 = 0 \quad \text{has no solution}
So no minimum — but this contradicts realistic models.
Reinterpreting: "cost per unit" is standard, and we must minimize $ P(x) = C(x)/x $. But if $ C(x) = 5000 + 120x - 0.5x^2 $, then for large $ x $, the $ -0.5x^2 $ dominates and cost per unit becomes negative — impossible.
Therefore, likely the quadratic coefficient is positive. Assume a typo, and the intended model is $ C(x) = 5000 + 120x + 0.5x^2 $. But the problem says $ -0.5x^2 $.
Alternatively, perhaps it's $ C(x) = 5000 + 120x + 0.5x^2 $? But it says $ -0.5x^2 $.

Alternatively, perhaps the function is \( C(x) = 5000 + 120x - 0.5x^2 \), and we want to minimize total cost, not per unit. But the problem says "cost per unit."

Related Articles

Trending Articles