Mathematics - Application of Derivative Question with Solution | TestHub

$h=r\sin \theta +r$

base $=BC=2r\cos \theta$

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

Area of $\mathrm{\Delta }ABC=\frac{1}{2}\left(BC\right).h$

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

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

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

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

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

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

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

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

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

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

$=\underset{\mathrm{positive}}{\underbrace{r^2\left[1+\mathrm{sin\theta }\right]}}\left[1-2\sin \theta \right]=0$

$\Rightarrow \theta =\frac{\pi }{6}$

$\Rightarrow$ $\mathrm{\Delta }$ is maximum where $\theta =\frac{\pi }{6}$

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