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Home » Volume is given by \( \textlength \times \textwidth \times \textheight = 240 \). - Blask

Volume is given by \( \textlength \times \textwidth \times \textheight = 240 \). - Blask

📅 March 10, 2026 👤 scraface

Mar 10, 2026

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📌 Simplify using $\sin^2 x + \cos^2 x = 1$ and $\csc^2 x = 1 + \cot^2 x$, $\sec^2 x = 1 + \tan^2 x$:

📌 + 2 + 2 + \cot^2 x + \tan^2 x = 5 + \cot^2 x + \tan^2 x.

📌 Let $t = \tan^2 x$, so $\cot^2 x = \frac{1}{t}$. The expression becomes:

📌 + t + \frac{1}{t}.

📌 By AM-GM, $t + \frac{1}{t} \geq 2$, with equality when $t = 1$ (i.e., $x = \frac{\pi}{4}$). Thus, the minimum is $7$, but we seek the maximum. As $x \to 0^+$ or $x \to \frac{\pi}{2}^-$, $\tan^2 x \to 0$ or $\infty$, so $t + \frac{1}{t} \to \infty$. However, the original expression is unbounded. Wait, this contradicts the problem's implication of a finite maximum. Re-examining:

📌 Actually, $\sin x + \csc x = \sin x + \frac{1}{\sin x} \geq 2$, and similarly for $\cos x + \sec x$. But squaring gives:

📌 But the sum is minimized at $8$ (when $x = \frac{\pi}{4}$), but maximum is unbounded. However, the problem likely seeks the minimum. Assuming a typo, if the question is to find the minimum:

📌 At $x = \frac{\pi}{4}$, $\sin x = \cos x = \frac{\sqrt{2}}{2}$, $\csc x = \sec x = \sqrt{2}$. Then:

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