Question

An open tank is to be constructed with a square base and vertical sides so as to contain a given quantity of water. Show that the expenses of lining with lead will be least, if depth is made half of width.

Answer 1

Let L be the length of the square base and h be the height/depth of the tank. 

Expenses of lining implies the cost for lining the entire inner surface area of the tank ; a base and four vertical sides. 

If we have minimum area to cover, we will have minimum costs incurred. 

Internal area of the tank = L² + 4Lh 

Volume of the tank = L²h = V 

Therefore,

We get two outcomes here.

L = 0,2h 

We discard L = 0, 

as it makes no sense. 

So,

L = 2h. 

Now to check whether a maxima or a minima exists

\(\frac{d^2A}{dL^2}\) = 2 + 4 x 2 x \(\frac{V}{L^3}\) > 0

Therefore a minima exists for all non zero values of L. 

Hence, 

for the tank lining costs to be minimum, h = \(\frac{L}{2}\).