<s>[INST] Given $S$ is maximum, find $\dfrac{h}{x}$ A question is given as such: Given $h=\dfrac{L(R+\sqrt{R^2-x^2})}{2R}$ and $S=\dfrac{4}{3}xh$, find $\dfrac{h}{x}$ in terms of $L$ and $R$ when $S$ is maximum.
By differentiation, I got that $R\sqrt{R^2-x^2}+R^2-2x^2=0$, which gives that $\dfrac{h}{x}=\dfrac{Lx}{R^2}$. On the other hand I can also get that $2h=\dfrac{L}{R}\dfrac{x^2}{\sqrt{R^2-x^2}}$, but neither of them gives anything close to $\dfrac{h}{x}$.
The final answer is $\dfrac{h}{x}=\dfrac{\sqrt{3}L}{2R}$.
 [/INST] If you know that $\frac{h}{x} = \frac{Lx}{R^2}$ and $2h = \frac{L}{R} \frac{x^2}{\sqrt{R^2-x^2}}$, you can solve for $x$ in terms of $L, R$. Multiplying the first equation by $x$ and dividing the second by $2$ yields $$\frac{Lx^2}{R^2} = h = \frac{L}{2R} \frac{x^2}{\sqrt{R^2-x^2}}$$
Multiplying both sides by $\frac{R}{Lx^2}$ makes it $$\frac{1}{R} = \frac{1}{2\sqrt{R^2-x^2}}$$
Then this means that $R^2-x^2 = \frac{R^2}{4} \to x^2 = \frac{3R^2}{4}$. Take the square root to get that $x = \frac{R\sqrt{3}}{2}$.
Using $\frac{h}{x} = \frac{Lx}{R^2}$, you can plug in $x = \frac{R\sqrt{3}}{2}$ to get $$\frac{h}{x} = \frac{L\frac{R\sqrt{3}}{2}}{R^2} = \frac{L\sqrt{3}}{2R}$$
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