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Home » -\frac27 \leq k \leq \frac977 - jntua results

-\frac27 \leq k \leq \frac977 - jntua results

📅 March 10, 2026 👤 scraface

Mar 10, 2026

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📌 I'(t) = \frac{1000(t^2 + 10) - 1000t(2t)}{(t^2 + 10)^2} = \frac{1000t^2 + 10000 - 2000t^2}{(t^2 + 10)^2} = \frac{-1000t^2 + 10000}{(t^2 + 10)^2}

📌 Set \( I'(t) = 0 \):

📌 -1000t^2 + 10000 = 0 \Rightarrow t^2 = 10 \Rightarrow t = \sqrt{10} \quad (\text{since } t \geq 0 \text{ for time})

📌 To confirm it’s a maximum, check the sign of \( I'(t) \): decreasing before \( t = \sqrt{10} \), increasing after? Actually, the numerator \( -1000t^2 + 10000 \) is positive for \( t < \sqrt{10} \) and negative for \( t > \sqrt{10} \), so \( I'(t) \) changes from positive to negative — maximum at \( t = \sqrt{10} \).

📌 Thus, the infection rate peaks at \( t = \sqrt{10} \) days.

📌 \(\boxed{\sqrt{10}}\)

📌 Question:** A venture capitalist is evaluating a cleantech startup’s cost model given by \( C(x) = 5000 + 120x - 0.5x^2 \), where \( C \) is the cost in dollars for producing \( x \) units. Find the production level \( x \) that minimizes cost per unit.

📌 The cost per unit is \( P(x) = \frac{C(x)}{x} = \frac{5000 + 120x - 0.5x^2}{x} = \frac{5000}{x} + 120 - 0.5x \), for \( x > 0 \).

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