The given sphere is of radius R. Let h be the height and r be the radius of the cylinder inscribed in the sphere.
Volume of cylinder
V = $\mathit{\pi }{\mathit{R}}^2\mathit{h}$ ...(1)
In right angled triangle ΔOBA

$\mathit{A}{\mathit{B}}^2+\mathit{O}{\mathit{B}}^2$ = $\mathit{O}{\mathit{A}}^2$
${\mathit{R}}^2+\frac{{\mathit{h}}^2}{4}$ = ${\mathit{r}}^2$
So, ${\mathit{R}}^2$ = ${\mathit{r}}^2-\frac{{\mathit{h}}^2}{4}$
Putting the value of ${\mathit{R}}^2$ in equation (1), we get
V = π $\left({\mathit{r}}^2-\frac{{\mathit{h}}^2}{4}\right)$ . h
V = π $\left({\mathit{r}}^2\mathit{h}-\frac{{\mathit{h}}^3}{4}\right)$ ... (3)
∴ $\frac{\mathit{d}\mathit{V}}{\mathit{d}\mathit{h}}$ = π $\left({\mathit{r}}^2-\frac{3{\mathit{h}}^2}{4}\right)$ ... (4)
For stationary point, $\frac{\mathit{d}\mathit{V}}{\mathit{d}\mathit{h}}$ = 0
π $\left({\mathit{r}}^2-\frac{3{\mathit{h}}^2}{4}\right)$ = 0
${\mathit{r}}^2$ = $\frac{3{\mathit{h}}^2}{4}$ , ⇒ ${\mathit{h}}^2$ = $\frac{4{\mathit{r}}^2}{3}$ , ⇒ h = $\frac{2\mathit{r}}{\sqrt{3}}$
Now , $\frac{{\mathit{d}}^2\mathit{V}}{\mathit{d}{\mathit{h}}^2}$ = π $\left(-\frac{6}{4}\mathit{h}\right)$
∴ ${\left[\frac{{\mathit{d}}^2\mathit{V}}{\mathit{d}{\mathit{h}}^2}\right]}_{\mathrm{at}\mathit{h}=\frac{2\mathit{r}}{\sqrt{3}}}$ = π $\left(-\frac{3}{2},\frac{2\mathit{r}}{\sqrt{3}}\right)$ < 0
∴ Volume is maximum at h = $\frac{2\mathit{r}}{\sqrt{3}}$
Maximum volume is = π $\left({\mathit{r}}^2\times \frac{2\mathit{r}}{\sqrt{3}}.\frac{1}{4}\times \frac{8{\mathit{r}}^3}{3\sqrt{3}}\right)$
= π $\left(\frac{2{\mathit{r}}^3}{\sqrt{3.}}\frac{2{\mathit{r}}^3}{3\sqrt{3}}\right)$
= π $\left(\frac{6{\mathit{r}}^3.2{\mathit{r}}^3}{3\sqrt{3}}\right)$
= $\frac{4\mathit{\pi }{\mathit{r}}^3}{4\sqrt{3}}$ cu. units
OR
Let l, b, and h denote the length breadth and depth of the open rectangular tank.
Given h = 2m
V = $8{\mathit{m}}^3$
i.e. 2/b = 8
⇒ lb = 4 or b = $\frac{4}{\mathit{l}}$
Surface area, S, of the open rectangular tank of depth 'h' = lb + 2 (l + b) × h
In this problem , b = $\frac{4}{\mathit{l}}$ , lb = 4 metre , h = 2 metre
∴ S = 4 + 2 (l + 4/l) × 2
⇒ S = 4 + 4 (l + 4/l)
For maxima or minima, differentiating with respect to l we get,
$\frac{\mathit{d}\mathit{S}}{\mathit{d}\mathit{l}}$ = 4 $\left(1-\frac{4}{{\mathit{l}}^2}\right)$
$\frac{\mathit{d}\mathit{S}}{\mathit{d}\mathit{l}}$ = 0 ⇒ l = 2m
l = 2m for minimum or maximum
Now, $\frac{{\mathit{d}}^2\mathit{S}}{\mathit{d}{\mathit{l}}^2}$ = $\frac{48}{{\mathit{l}}^3}$ > 0 for all l
So l = 2m is a point of minima and minimum surface area is
S = lb + 2 (l + b) × h
= 4 + 2 × 8 = 4 + 16 = 20 square meters
Base Area = 4 square metres; Lateral surface area = 16 square metres
cost = 4 × 70 + 16 × 45
= 280 + 720 = Rs. 1000