Let #x# be the length of the shorter edge of the bottom of the box so that #7x# is the length of the longer edge. Let #h# be the height of the box. The volume of the box is #V=x^2h# and the surface area, excluding the top, is #S=7x^2+2xh+14xh=7x^2+16xh#.
Since the volume #V# is fixed, we can solve for #h# in terms of #x# to get #h=V/(x^2)# and the substitute this into the equation for the surface area to get #S# as a function of #x#: #S=f(x)=7x^2+(16V)/x# for #x>0#.
We want to minimize #S=f(x)# for #x>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}#.
