A square tank of capacity 250 cubic metres has to be dug out. The cost of the land is ₹50 per square metre. The cost of digging increases with the depth and for the whole tank it is $₹400 × h^2$ where h metres is the depth of the tank. What should be the dimensions of the tank so that the cost would be minimum?

Side length = 10 m, Depth = 2.5 m

The correct answer is Option (2) → Side length = 10 m, Depth = 2.5 m

Let x metres be the side of the square tank, then the area of the base of the tank = $x^2$ sq. metres.

∴ Cost of land = $₹(50 × x^2)=₹50x^2$.

Cost of digging the tank = $₹400\, h^2$.

∴ Total cost of the tank = $₹(50x^2 + 400\, h^2)$.

Given capacity of the tank = 250 cubic metres,

$∴ x^2 × h = 250⇒h=\frac{250}{x^2}$   ...(i)

Let ₹C be the total cost of the tank, then

$C = 50x^2 + 400h^2 = 50x^2 + 400 ×\left(\frac{250}{x^2}\right)$   (using (i))

$=50x^2+\frac{25×10^6}{x^4}$

Diff. it w.r.t. x, we get

$\frac{dC}{dx}= 50.2x + 25 × 106.\left(\frac{-4}{x^5}\right)=100x -\frac{10^8}{x^5}$

And $\frac{d^2C}{dx^2}=100-10^8.\left(\frac{-5}{x^6}\right)=100+\frac{5 × 10^8}{x^6}$.

Now $\frac{dC}{dx}= 0 ⇒ 100 x-\frac{10^8}{x^5}=0⇒x^6=10^6⇒x=10$

Also $\left(\frac{d^2C}{dx^2}\right)_{x=10}=100+\frac{5 × 10^8}{x^6}=100+500=600>0$

⇒ C is minimum when $x = 10$.

When $x = 10$, from (i), $h =\frac{250}{100}= 2.5$

Hence, the dimensions of the tank are - side of the square base is 10 metres and depth of rank = 2.5 metres.