A tank with rectangular base and rectangular sides,

Solution:

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

Then, we have height (h) = 2 m

Volume of the tank = 8m3

Volume of the tank = l × b × h

∴ 8 = l × b × 2

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

Now, area of the base = lb = 4

Area of the 4 walls (A) = 2h (l + b)

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

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

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

$\Rightarrow 1-\frac{4}{l^{2}}=0$

$\Rightarrow l^{2}=4$

$\Rightarrow l=\pm 2$

However, the length cannot be negative.

Therefore, we have l = 4.

$\therefore b=\frac{4}{l}=\frac{4}{2}=2$

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

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

Thus, by second derivative test, the area is the minimum when l = 2.

We have l = b = h = 2.

∴Cost of building the base = Rs 70 × (lb) = Rs 70 (4) = Rs 280

Cost of building the walls = Rs 2h (l + b) × 45 = Rs 90 (2) (2 + 2)

= Rs 8 (90) = Rs 720

Required total cost = Rs (280 + 720) = Rs 1000

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