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

Step-by-step solution

Let x be the length of the side of square bottom, h be the height, V be the volume and A be the surface area of open tank. Then, aligned & V =x^2 ~h =4000...(i) & ~A =x^2+4 x ~h ...(ii) aligned From (i), h = 4000 x^2 Substituting the value of h in (ii), we get aligned & A =x^2+ 16000 x & dA ~d x =2 x- 16000 x^2 aligned A is minimum, if d A d x =0 aligned & 2 x- 16000 x^2 =0 & x^3=8000 & x=20 aligned Now, d ^2 ~A ~d x^2 =2+ 32000 x^3 ( d ^2 ~A ~d x^2 )_ x=20 =6 0 A is minimum when x=20 ~cm . h = 4000 x^2 = 4000 400