Let the length and breadth of the tank be $x$ metre and $y$ metre. The depth of it is 2 metre
$\therefore$ Volume of tank $=2 \times \mathrm{x} \times \mathrm{y}=2 \mathrm{xy}=8$
$\Rightarrow \text { xy }=4 \quad \ldots(i)$
$\therefore$ Area of base $=x y$, Area of sides $=4(x+y)$ Cost of construction
$=₹[70 \mathrm{xy}+180(\mathrm{x}+\mathrm{y})] \quad \ldots(ii)$
Putting value of $y$ in (ii) from (i), $y=\frac{4}{x}$,
$\therefore \quad C=70 \times 4+180\left(x+\frac{4}{x}\right)$
or $\frac{\mathrm{dC}}{\mathrm{dx}}=180\left(\frac{\mathrm{x}^2-4}{\mathrm{x}^2}\right)$,
Formaximum or minimum $\frac{\mathrm{dC}}{\mathrm{dx}}=0 \Rightarrow \mathrm{x}=\pm 2$
$\frac{\mathrm{dC}}{\mathrm{dx}}$ changes sign from $-$ ve to $+$ ve at $\mathrm{x}=2$
$\therefore \quad$ C is minimum at $\mathrm{x}=2, \therefore \mathrm{x}=2$ and $\mathrm{y}=2$,
$\therefore \quad$ Thus tank is a cube of side $2 \mathrm{~m}$
$\therefore \quad$ Least cost of construction $=₹ 1000$.