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Home » $ \frac126 \approx 0.0384615 $ - 500apps

$ \frac126 \approx 0.0384615 $ - 500apps

📅 March 13, 2026 👤 scraface

Mar 13, 2026

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📌 Let’s denote the denominator as $ D(t) = 1 + 9e^{-0.5t} $, so:

📌 P(t) = \frac{1000}{D(t)} \Rightarrow P'(t) = -1000 \cdot \frac{D'(t)}{[D(t)]^2}.

📌 Now compute $ D'(t) $:

📌 D(t) = 1 + 9e^{-0.5t} \Rightarrow D'(t) = 9 \cdot (-0.5)e^{-0.5t} = -4.5e^{-0.5t}.

📌 P'(t) = -1000 \cdot \frac{-4.5e^{-0.5t}}{(1 + 9e^{-0.5t})^2} = \frac{4500e^{-0.5t}}{(1 + 9e^{-0.5t})^2}.

📌 To find the maximum of $ P'(t) $, we define:

📌 f(t) = \frac{e^{-0.5t}}{(1 + 9e^{-0.5t})^2}.

📌 Let $ u = e^{-0.5t} $, so $ f(t) = \frac{u}{(1 + 9u)^2} $. Now find the maximum of this function with respect to $ u > 0 $.

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