Mathematics - Application of Derivative Question with Solution | TestHub

Sphere of radius $r=3\mathrm{cm}$ Let $b,h$ be base radius and height of cone respectively. So, volume of cone $=\frac{1}{2}\pi b^{2}h$
In right angled $△ABC$ by Pythagoras theorem $$\begin{array}{rl}& (h-r{)}^2+b^2=r^2\\ \Rightarrow & b^2=r^2-(h-r{)}^2=r^2-\left(h^2-2hr+{\mathrm{r}}^2\right)=2hr-h^2\\ \therefore & \text{ Volume }(v)\frac{1}{3}\pi h\left[2hr-h^2\right]=\frac{1}{3}\left[2h{r}^2-h^3\right]\\ & \frac{dv}{dh}=\frac{1}{3}\left[4hr-3h^2\right]=0\Rightarrow h(4r-3h)=0\\ & \frac{d^{2}v}{d{h}^2}=\frac{1}{3}[4r-6h]\end{array}$$ At $h=\frac{4r}{3},\frac{d^{2}v}{d{h}^2}=\frac{1}{3}\left[4r-\frac{4r}{3}\times 6\right]=\frac{1}{3}[4r-8r<]0$ $\Rightarrow$ maximum volume ocurs at $h$ $$=\frac{4r}{3}=\frac{4}{3}\times 3=4\mathrm{cm}$$ As from (1), $$(h-r{)}^2+b^2=r^2$$ $$\Rightarrow b^2=2hr-h^2=2\cdot \frac{4r}{3}r-\frac{16r^2}{9}$$ $$\begin{array}{rl}& =\frac{8r^2}{3}-\frac{16r^2}{9}\\ & =\frac{(24-16)r^2}{9}=\frac{8r^2}{9}\\ & \Rightarrow b=\frac{2\sqrt{2}}{3}r=2\sqrt{2}\mathrm{cm}\end{array}$$ Therefore curved surface area $$\begin{array}{rl}& =\pi bl=\pi b\sqrt{h^2+r^2}\\ & =\pi 2\sqrt{2}\sqrt{4^2+8}=8\sqrt{3}\pi {\mathrm{cm}}^2\end{array}$$