The triangle of maximum area that can be inscribed in a given...

$h = rsin\theta + r$

base $= BC = 2rcos\theta$

$\theta \in \left[0, \dfrac{\pi}{2}\right)$

Area of $\delta ABC = \dfrac{1}{2}\left(BC\right) . h$

$\Delta = \dfrac{1}{2}\left(2rcos\theta\right) . \left(rsin\theta + r\right)$

$= {r}^{2}\left(cos\theta\right) . \left(1 + sin\theta\right)$

$\dfrac{d\Delta}{d\theta} = {r}^{2}\left[cos\theta\dfrac{d}{d\theta}\left(1 + sin\theta\right) + \left(1 + sin\theta\right)\dfrac{d}{d\theta}\left(cos\theta\right)\right]$

$\dfrac{d\Delta}{d\theta} = {r}^{2}\left[cos\theta\left(0 + cos\theta\right) + \left(1 + sin\theta\right)\left(- sin\theta\right)\right]$

$\dfrac{d\Delta}{d\theta} = {r}^{2}\left[\cos^{2}\theta - sin\theta - \sin^{2}\theta\right]$

$\dfrac{d\Delta}{d\theta} = {r}^{2}\left[1 - \sin^{2}\theta - sin\theta - \sin^{2}\theta\right]$

$= {r}^{2}\left[1 - sin\theta - 2\sin^{2}\theta\right]$

$= {r}^{2}\left[- 2\sin^{2}\theta - sin\theta + 1\right]$

$= {r}^{2}\left[- 2\sin^{2}\theta - 2sin\theta + sin\theta + 1\right]$

$= {r}^{2}\left[- 2sin\theta\left(sin\theta + 1\right) + 1\left(sin\theta + 1\right)\right]$

$= \mathop{\mathop{{r}^{2}\left[1 + sin \theta\right]}_{⏟}}_{positive}\left[1 - 2sin\theta\right] = 0$

$⇒ \theta = \dfrac{\pi}{6}$

$⇒$ $\Delta$ is maximum where $\theta = \dfrac{\pi}{6}$

{\Delta}_\max = \dfrac{3\sqrt{3}}{4}{r}^{2} = area of equilateral $\Delta$ with $BC = \sqrt{3}r$.