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.
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
Therefore a minima exists for all non zero values of L.
Hence for the tank lining costs to be minimum, h = L/2.