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Home » Temps pour la premiÃ¨re partie : \( rac15050 = 3 \) heures. - Esdistancia

Temps pour la premiÃ¨re partie : \( rac15050 = 3 \) heures. - Esdistancia

📅 March 17, 2026 👤 scraface

Mar 17, 2026

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10x - 5x^2 = 0 \quad \Rightarrow \quad 5x(2 - x) = 0.
Thus, \( x = 0 \) or \( x = 2 \).
To determine which value maximizes \( C(x) \), evaluate the second der...
C''(x) = rac{d}{dx}(e^{-x}(10x - 5x^2)) = e^{-x}(-10x + 5x^2) + e^{-x...
C''(x) = e^{-x}(-10x + 5x^2 + 10 - 10x) = e^{-x}(5x^2 - 20x + 10).
Evaluate \( C''(x) \) at the critical points:
\( x = 0 \):
C''(0) = e^0(5(0)^2 - 20(0) + 10) = 10 > 0 \quad ( ext{local minimum})...
C''(2) = e^{-2}(5(2)^2 - 20(2) + 10) = e^{-2}(20 - 40 + 10) = e^{-2}(-...
Therefore, the concentration is maximized at \( x = 2 \).

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