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Home » Revolutionize Your Networking: The Ultimate Calling Card That Gets Results - Leaselab

Revolutionize Your Networking: The Ultimate Calling Card That Gets Results - Leaselab

📅 February 27, 2026 👤 scraface

Feb 27, 2026

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📌 Set \( P'(x) = 0 \): no solution. So minimum does not exist — function decreases for large \( x \)? No, \( -0.5x \) dominates, so \( P(x) \to -\infty \), impossible.

📌 This suggests the model is invalid, but for math olympiad, likely the function is \( C(x) = 5000 + 120x + 0.5x^2 \). But it says \( -0.5x^2 \).

📌 Wait — perhaps it's \( C(x) = 5000 + 120x - 0.5x^2 \), but \( x \) has an upper limit. But no.

📌 Alternatively, the term \( -0.5x^2 \) may be a typo, and it should be \( +0.5x^2 \). Let's assume that, as it's common in real models.

📌 So assume \( C(x) = 5000 + 120x + 0.5x^2 \). Then average cost:

📌 P(x) = \frac{5000}{x} + 120 + 0.5x

📌 Now take derivative:

📌 P'(x) = -\frac{5000}{x^2} + 0.5

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× 8.75373 = 105.04476 ≈ 105.04 N
But since the increases are additive and the model assumes multiplicat...
Total = 12 + 12(0.15) + 12(0.15)(1.15) + 12(0.15)(1.15)^2 + ... for 6...
This is geometric series with a = 12, d = 12×0.15 = 1.8 (but actually...
Let F = sum of forces: F = 12 + 12(1.15) + 12(1.15)^2 + 12(1.15)^3 + 1...
F = 12 [ (1.15^6 - 1) / (1.15 - 1) ] = 12 [ (2.31306 - 1) / 0.15 ] = 1...
Rounded to nearest hundredth: 105.04
But for math olympiad precision, use exact calculation:
1.15^6 = (23/20)^6 ≈ 2.31306, same as above.
So final force ≈ 105.04 N

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