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Home » The Untold Betrayal of Sulla: How One Man Rocked Ancient History—Crazy Stories! - Appcentric

The Untold Betrayal of Sulla: How One Man Rocked Ancient History—Crazy Stories! - Appcentric

📅 February 27, 2026 👤 scraface

Feb 27, 2026

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📌 Wait — likely the model is \( C(x) = 5000 + 120x + 0.5x^2 \) for increasing marginal cost, but given is \( -0.5x^2 \), which causes average cost to eventually decrease — not realistic.

📌 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.

📌 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 \( -0.5x^2 \).

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Complete the square for the \( x \)-terms:
9(x^2 - 4x) = 9((x-2)^2 - 4) = 9(x-2)^2 - 36
Complete the square for the \( y \)-terms:
4(y^2 + 4y) = 4((y+2)^2 - 4) = 4(y+2)^2 - 16
Substitute these back into the equation:
9(x-2)^2 - 36 + 4(y+2)^2 - 16 = -36
9(x-2)^2 + 4(y+2)^2 - 52 = -36
Add 52 to both sides:
9(x-2)^2 + 4(y+2)^2 = 16
Divide the entire equation by 16 to get the standard form:

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