[SOLVED] Minimizing the Surface Area 1. The problem statement, all variables and given/known data A box has a bottom with one edge 8 times as long as the other. If the box has no top and the volume is fixed at V, what dimensions minimize the surface area? 2. Relevant equations V = lwh SA (with no top) = lw + 2lh + 2wh 3. The attempt at a solution l = x w = 8x h = V/(8x^2) Finding an equation for the surface area. SA = lw + 2lh + 2wh SA = x(8x) + 2x(V/(8x^2)) + 2(8x)(V/(8x^2)) SA = 8x^2 + V/(4x) + 2V/x Finding the derivative of the equation in order to set it equal to zero to find the critical points, so the minimum can be found. (d SA)/(d x) = 16x - V/(4x^2) - 2V/(x^2) (d SA)/(d x) = (64x^3 - 9V) / (4x^2) (d SA)/(d x) = 0 (64x^3 - 9V) / (4x^2) = 0 (64x^3 - 9V) = 0 64x^3 = 9V x^3 = (9V)/64 x = ((9V)/64)^(1/3) Plugging the solution into the equations for the dimensions. l = x = ((9V)/64)^(1/3) w = 8x = 8((9V)/64)^(1/3) h = V/(8(((9V)/64)^(1/3))^2) I am unsure if I did the right steps in order to find the solution. Also, I am not very confident in the work I did for each step.