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V = \frac16 \left| \det \beginbmatrix - Crosslake

📅 March 1, 2026 👤 scraface

Mar 01, 2026

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📌 Innere Abmessungen: \( (20 - 2x) \times (15 - 2x) = 240 \).

📌 Erweitern: \( 300 - 40x - 30x + 4x^2 = 240 \) → \( 4x^2 - 70x + 300 = 240 \) → \( 4x^2 - 70x + 60 = 0 \).

📌 Dividiere durch 2: \( 2x^2 - 35x + 30 = 0 \).

📌 Verwende quadratische Formel: \( x = \frac{35 \pm \sqrt{(-35)^2 - 4 \cdot 2 \cdot 30}}{2 \cdot 2} = \frac{35 \pm \sqrt{1225 - 240}}{4} = \frac{35 \pm \sqrt{985}}{4} \).

📌 Approximiere \( \sqrt{985} \approx 31.4 \) → \( x \approx \frac{35 - 31.4}{4} = \frac{3.6}{4} = 0.9 \).

📌 Prüfe: bei \( x = 0.5 \): Fläche = \( 19 \times 14 = 266 > 240 \); bei \( x = 1 \): \( 18 \times 13 = 234 < 240 \); bei \( x = 0.8 \): \( 19.4 \times 13.4 = 260.36 \); besser exakt:

📌 Tatsächliche Lösung durch Faktorversuch: versuche \( x = 0.6 \): \( 19.8 \times 13.8 = 273.24 \) — zu hoch.

📌 Verwende korrekt: \( 2x^2 - 35x + 30 = 0 \) → \( x = \frac{35 - \sqrt{1225 - 240}}{4} = \frac{35 - \sqrt{985}}{4} \).

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