Among all closed rectangular boxes of volume 27 cm3, what is the smallest surface area?
Answers
For reasons of symmetry it will be a cubic box 3 cm on each edge.
If you must prove this, I'd do it in two steps.
(1) show that if two edges are equal, the best solution is obtained when the 3rd edge equals the other two, and
(2) consider the case where all three edges are unequal.
(1) Let x be one of two equal edges.
V = 27 = hx^2 and
A = 2x^2 + 4hx
From the "V" equation, rewrite A:
A = 2x^2 + 4(27/x^2)x = 2x^2 + 108/x
dA/dx = 4x - 108/x^2
When dA/dx = 0, you have
4x = 108/x^2
x^3 = 27 => x = 3
Then h = 27/x^2 = 3 also.
(2) The harder case where the edges are x,y, and h.
V = xyh = 27
A = 2xy + 2yh + 2xh
A = 2xy + 2y(27/xy) + 2x(27/xy)
A = 2xy + 54/x + 54/y
Now take the PARTIAL derivatives of A
(sorry I can't type those backwards-6 things):
dA/dx = 2y - 54/x^2
dA/dy = 2x - 54/y^2
We require both of these to be zero, so
2y = 54/x^2
y = 27/x^2
dA/dy = 2x - 54/[(27/x^2)^2]
dA/dy = 2x - 2x^4/27
Setting this to zero you get
54x = 2x^4
Insisting that x not be zero, you get
27 = x^3 => x = 3; then y = 27/x^2 = 3
and h = V/(xy) = 27/9 = 3.
Part (1) could be eliminated, it's a just a way of getting at the thought process.