A tank with rectangular base and rectangular sides, open at the top is to constructed so that its depth 2 m and volume is 8 m³. If building of tank costs Rs. 70 per square metre for the base and Rs. 45 per square metre for the sides, what is the cost of least expensive tank?

Assume that l, b and h are the length, width and height of the tank respectively,
Given,

Height, h=2m

Volume of the tank=8 ${\mathrm{m}}^3$

Volume of the tank = $l\times b\times h$

$$\begin{gathered} \therefore \mathrm{l}\times \mathrm{b}\times 2=8 \\ \Rightarrow lb=4 \\ \Rightarrow b=\frac{4}{l} \end{gathered}$$

Area of the base =lb=4 ${\mathrm{m}}^2$

Area of the 4 walls, A=2h(l+b)

$$\begin{gathered} \therefore A=4\left(l+\frac{A}{l}\right) \\ \Rightarrow \frac{dA}{dl}=4\left(1-\frac{4}{l^2}\right) \end{gathered}$$

For maximum or minimum values of A, we must have 

$$\begin{gathered} \frac{dA}{dl}=0 \\ \Rightarrow 4\left(1-\frac{4}{l^2}\right)=0 \\ \Rightarrow l=\pm 2 \end{gathered}$$

We know that the length cannot be negative, so l=2m.

$\therefore \mathrm{b}=\frac{4}{2}=2\mathrm{m}$

Now,

$$\begin{gathered} \frac{{\mathrm{d}}^2\mathrm{A}}{{\mathrm{dl}}^2}=\frac{32}{{\mathrm{l}}^2} \\ \mathrm{At} l=2 : \\ \frac{{\mathrm{d}}^2\mathrm{A}}{d{l}^2}=\frac{32}{8}=4>0 \end{gathered}$$

Thus, the area is the minimum when l=2 m

We have

l=b=h=2 m

$\therefore$ Cost of building the base $=\mathrm{Rs} 70\times \left(\mathrm{lb}\right)=\mathrm{Rs}70\mathrm{v}4=\mathrm{Rs} 280$

Cost of building the walls $=\mathrm{Rs} 2\mathrm{h}(\mathrm{l}+\mathrm{b})\times 45=\mathrm{Rs}90\left(2\right) (2+2\mathrm{Rs} 8(90)=\mathrm{Rs} 720$

Total cost $=(280+720)=\mathrm{Rs} 1000$

Therefore, the cost of the least expensive tank is Rs. 1000.