If a box has a square base its volume will be:
V=hb^2 where h is the height...
h=V/b^2 we are told that V=125 so
h=125/b^2 now for the surface area, which consists of the two bases for a total of 2b^2. It will also have four sides with a total area of 4(bh)=4bh so
A=2b^2+4bh, using h found above in this gives us:
A=2b^2+4b(V/b^2)
A=2b^2+4V/b
A=(2b^3+4V)/b, then taking the derivatives we can find the velocity of the area function.
dA/db=(6b^3-2b^3-4V)/b^2
dA/db=(4b^3-4V)/b^2
dA/db=0 when 4b^3-4V=0
b^3=V
b=V^(1/3), since V=125
b=5in, and since h=V/b^2
h=125/25=5in
So the dimensions that will minimize the amount of material used to enclose a volume of 125in^2 is a 5in cube.
h=b=5in
