A rectangular box open at the top is to have a volume 32cc. Find the dimensions of the box , that requires the least material for its construction.
Answers
Answer:
It is an easy homework assignment which you should try yourself.
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.
Answer:
If the box is square with side length x, the surface area is
a = xy+2yz+2xz
= x^2 + 2z(2x)
= x^2 + 4x(32/(x^2))
= x^2 + 128/x
This will be minimized when the derivative with respect to x is zero.
2x - 128/x^2 = 0
2x^3 - 128 = 0
x^3 = 64
x = 4
A square box that is 4 ft by 4 ft and 2 ft deep will have minimum surface area.