Question

An open tank with a square base and vertical sides is to be constructed from a metal sheet so as to hold a given quantity of water. Show that the cost of the material will be least when depth of the tank is half of its width.

Answer 1

Let x be the side of the square base h be the depth
Then, `V = x^(2)h`
Let the cost per square metre be Rs p. Then,
`C = (x^(2) + 4xh) p rArr C = (x^(2) + 4x xx (V)/(x^(2)))p rArr C = (x^(2) + (4V)/(x))p`
`:. (dC)/(dx) = (2x - (4V)/(x^(2)))p and (d^(2)C)/(dx^(2)) = (2 + (8V)/(x^(3)))p`
Now, `(dC)/(dx) = 0 hArr (2x - (4V)/(x^(2))) = 0 hArr x = (2V)^(1//3)`
And, `[(d^(2)C)/(dx^(2))]_(x = (2V)^(1//3)) = p (2 + (8V)/(2V)) = 6p gt 0`
`:.` C is minimum, when `x = (2V)^(1//3)`
Then, `h = (V)/(x^(2)) = (1)/(2) (2V)^(1//3) = (1)/(2)x`
Hence, for minimum cost, the depth of the tank should be equal to half of the side of its square base.