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Home » Set 8w = 48, so \( w = 48/8 = 6 \). - Tacotoon

Set 8w = 48, so \( w = 48/8 = 6 \). - Tacotoon

📅 March 10, 2026 👤 scraface

Mar 10, 2026

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📌 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."

📌 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 \( -0.5x^2 \).

📌 Given the instructions, we proceed with the **given** function, but realize: to minimize \( P(x) = \frac{5000}{x} + 120 - 0.5x \), take derivative:

📌 Set to zero: no solution. So no critical point — minimum occurs at boundary? But unbounded below? Not physical.

📌 Wait — better: perhaps the cost function is \( C(x) = 5000 + 120x - 0.5x^2 \), and we minimize **total cost**, but the phrase is "cost per unit".

📌 Given the confusion, let's assume the function is meant to be minimized as total cost, but the problem says "cost per unit".

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