Let the length, breadth and depth of the open tank be x, x and y, respectively. Length and breadth are same because given tank has a square base. Again, let V denotes its volume and S denotes its surface area. Now, given that
V = x2y ......(i)
Also, we know that the total surface area of the open tank is given by
S = x2 + 4xy ....(ii)
So, depth of tank is half of its width.
or S is minimum.
Hence, total surface area of the tank is least, when depth is half of its width