Step 1 :The problem is asking for the dimensions of a rectangular tank with a square base that has the minimum surface area. The volume of the tank is given as 2916 cubic feet.
Step 2 :The volume of a rectangular tank with a square base is given by the formula \(V = x^2 * h\), where \(x\) is the length of the side of the square base and \(h\) is the height of the tank.
Step 3 :The surface area of the tank (excluding the open top) is given by the formula \(A = x^2 + 4*x*h\).
Step 4 :We want to minimize the surface area \(A\) while keeping the volume \(V\) constant at 2916 cubic feet.
Step 5 :This is a problem of optimization with constraints. We can solve it using calculus.
Step 6 :First, we can express \(h\) in terms of \(x\) using the volume equation: \(h = V / x^2 = 2916 / x^2\).
Step 7 :Then we can substitute \(h\) into the surface area equation to get \(A\) in terms of \(x\) only: \(A = x^2 + 4*x*(2916 / x^2) = x^2 + 11664 / x\).
Step 8 :We can find the minimum of \(A\) by taking the derivative of \(A\) with respect to \(x\), setting it equal to zero, and solving for \(x\).
Step 9 :The critical points are 18, -9 - 9*sqrt(3)*I, and -9 + 9*sqrt(3)*I. The negative and complex solutions are not meaningful in this context, so we discard them. The only meaningful solution is \(x = 18\).
Step 10 :We can substitute \(x = 18\) into the equation for \(h\) to find the height of the tank.
Step 11 :The height of the tank is 9 feet.
Step 12 :So, the dimensions of the tank that minimize the surface area are a base of 18 feet by 18 feet and a height of 9 feet.
Step 13 :Final Answer: The dimensions of the tank with minimum surface area are \(\boxed{18, 18, 9}\) feet.