A tank with rectangular base and rectangular sides,

Solution:

Let $l, b$ and $h$ be the length, breadth and height of the tank, respectively.

Height, $h=2 \mathrm{~m}$

Volume of the tank $=8 \mathrm{~m}^{3}$

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

$\therefore l \times b \times 2=8$

$\Rightarrow l b=4$

$\Rightarrow b=\frac{4}{l}$

Area of the base $=1 b=4 \mathrm{~m}^{2}$

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

$\therefore A=4\left(l+\frac{4}{l}\right)$

$\Rightarrow \frac{d A}{d l}=4\left(1-\frac{4}{l^{2}}\right)$

For maximum or minimum values of $A$, we must have

$\frac{d A}{d l}=0$

$\Rightarrow 4\left(1-\frac{4}{1^{2}}\right)=0$

$\Rightarrow l=\pm 2$

However, the length cannot be negative.

Thus,

$I=2 \mathrm{~m}$

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

Now,

$\frac{d^{2} A}{d l^{2}}=\frac{32}{l^{3}}$

$\mathrm{At} 1=2:$

$\frac{d^{2} A}{d l^{2}}=\frac{32}{8}=4>0$

Thus, the area is the minimum when $/=2 \mathrm{~m}$

We have

$l=b=h=2 \mathrm{~m}$

$\therefore$ Cost of building the base $=$ Rs $70 \times(l b)=$ Rs $70 \times 4=$ Rs 280

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

Total cost $=$ Rs $(280+720)=$ Rs 1000

Hence, the total cost of the tank will be Rs 1000 .

Disclaimer: The solution given in the book is incorrect. The solution here is created according to the question given in the book.