A rectangular box open at the top is to have volume of 32cubic feet. Find dimension of the box requiring least material for it's construction?

Dear Student

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.

Hence the box has length and breadth as4ft and a height of2ft.

Thanks