If the length, breadth and height of the box is denoted by a, b and h respectively, then V=a×b×h =32, and so h=32/ab. Now we have to maximize the surface area (lateral and the bottom) A = (2ah+2bh)+ab =2h(a+b)+ab = [64(a+b)/ab]+ab =64[(1/b)+(1/a)]+ab.
We treat A as a function of the variables and b and equating its partial derivatives with respect to a and b to 0. This gives {-64/(a^2)}+b=0, which means b=64/a^2. Since A(a,b) is symmetric in a and b, partial differentiation with respect to b gives a=64/b^2, ==>a=64[(a^2)/64}^2 =(a^4)/64. From this we get a=0 or a^3=64, which has the only real solution a=4. From the above relations or by symmetry, we get b=0 or b=4. For a=0 or b=0, the value of V is 0 and so are inadmissible. For a=4=b, we get h=32/ab =32/16 = 2.
Therefore the box has length and breadth as 4 ft each and a height of 2 ft.