An open tank with a square bottom is to contain 4000 cubic cm . of liquid. The dimensions of the tank so that the surface area of the tank is minimum, is

Step-by-step solution

The open tank has a square base with side length x and height h , with volume constraint x^2 h = 4000 , yielding h = 4000 x^2 . The surface area consists of the base and four sides: A = x^2 + 4xh . Substituting for h gives A(x) = x^2 + 16000 x . Differentiating to minimize area: A'(x) = 2x - 16000 x^2 . Setting A'(x) = 0 yields 2x = 16000 x^2 , so 2x^3 = 16000 , and x^3 = 8000 . Thus x = 20 cm. Substituting back, h = 4000 20^2 = 10 cm. The second derivative A''(x) = 2 + 32000 x^3 is positive at x = 20 , confirming