Let xx be the length of the shorter edge of the bottom of the box so that 7x7x is the length of the longer edge. Let hh be the height of the box. The volume of the box is V=x^2hV=x2h and the surface area, excluding the top, is S=7x^2+2xh+14xh=7x^2+16xhS=7x2+2xh+14xh=7x2+16xh.
Since the volume VV is fixed, we can solve for hh in terms of xx to get h=V/(x^2)h=Vx2 and the substitute this into the equation for the surface area to get SS as a function of xx: S=f(x)=7x^2+(16V)/xS=f(x)=7x2+16Vx for x>0x>0.
We want to minimize S=f(x)S=f(x) for x>0x>0, so take its derivative to get (dS)/dx=f'(x)=14x-(16V)/x^2. Now solve for any positive critical points by setting this equal to zero to get the equation 14x=(16V)/x^2 or x^3=(8V)/7 and x=(2V^(1/3))/7^(1/3)\approx 1.04552V^(1/3).
That this is the unique point that minimizes the surface area follows since (d^2S)/(dx^2)=f''(x)=14+(32V)/(x^3)>0 for all x>0, making the graph of S=f(x) concave up for all x>0.
The dimensions that minimize the volume are therefore x=(2V^(1/3))/7^(1/3)\approx 1.04552V^{1/3}, 7x=2*7^{2/3}V^{1/3}\approx 7.31861V^{1/3}, and h=V/(x^2)=(7^{2/3})/4 V^{1/3}\approx 0.91483 V^{1/3} and the minimum surface area is S=f((2V^(1/3))/7^(1/3))=12*7^{1/3}V^{2/3}\approx 22.95517V^{2/3}.
