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Home » Try \( x = 1 \) again â not zero. - NBX Soluciones

Try \( x = 1 \) again â not zero. - NBX Soluciones

📅 May 3, 2026 👤 scraface

May 03, 2026

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📌 r^2 = \frac{100}{\sin(2\theta)}.

📌 Since $r^2 \geq 0$, we must have $\sin(2\theta) > 0$. The maximum value of $r^2$ occurs when $\sin(2\theta)$ is minimized (but still positive). The minimum positive value of $\sin(2\theta)$ is approaching 0, but we must consider the domain where the expression is defined and real.

📌 However, to **maximize $r$**, we must **minimize the denominator** $\sin(2\theta)$, but not go below 0. So we consider the **maximum of $r^2$** under the constraint that $\sin(2\theta) > 0$. The maximum occurs when $\sin(2\theta)$ is at its **minimum positive value**, but that would make $r^2$ large — but we must find the **actual maximum possible finite value**.

📌 Actually, the **maximum finite value of $r$** occurs when $\sin(2\theta)$ is at its **maximum**, because $r^2 = \frac{100}{\sin(2\theta)}$, and $\sin(2\theta) \leq 1$.

📌 r^2 \leq \frac{100}{1} = 100 \Rightarrow r \leq 10.

📌 But we must check if this maximum is **attainable**. That is, does there exist a $\theta$ such that $\sin(2\theta) = 1$ and $r = 10$?

📌 Set $\sin(2\theta) = 1 \Rightarrow 2\theta = \frac{\pi}{2} + 2k\pi \Rightarrow \theta = \frac{\pi}{4} + k\pi$.

📌 Try $\theta = \frac{\pi}{4}$:

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