differential calculus
posted by jr on .
a pc of wire 10feet long is cut into 2 pcs. one piece is bent into the shape of a circle and the other into the shape of a square. how should the wire be cut so that the combined area is as large as possible?

Intuitively, the circle has a bigger area for the same circumference, so we try to make a bigger circle.
Also, since the area increases as the square of the circumference, we want to make 1 single shape as opposed to two.
So intuitive answer is to make the smallest square possible (zero if allowed).
Let's check our intuition.
Let x=length to make the circle
Radius of circle, r
=x/(2π)
Area of circle
=πr²
=x²/(4π)
For the square,
perimeter=10x
side length, s
=(10x)/4
Area of square
=((10x)/4)²
Total area, A(x)
=x^2/(4*π)+((10x)/4)²
dA(x)/dx=x/(2π)(10x)/8
For maximum, equate dA(x)/dx to zero, and solve for x.(=4.399)
Check if the answer is a maximum or minimum by graphing, or calculate d^{2}A(x)/dx².
If d^{2}A(x)/dx².>0, it is a minimum, and thus has no use for us.
In that case, we calculate A(0), A(4.399) and A(10) to find the areas at the critical points and choose the largest one. 
This is a very common question in introductory Calculus, and each time I have seen it, it asked for a MINIMUM area , and as MathMate noted, we would make the circle as large as possible.
Taking it with the usual minimum area request,
let the radius of the circle be r
let the side of the square be x
then 2πr + 4x = 10
x = (5  πr)/2
A = πr^2 + x^2 = πr^2 + (1/4) (25  10πr + π^2 r^2)
dA/dr = 2πr + (1/4)(10 +2π^2 r) = 0 for a min of A
dividing by π and solving for r,
r = 5/(4+π) = appr .7001
x = 1.4002
since 4x is needed for the square, the wire should be cut
appr 5.6 cm for the square and 4.4 cm for the circle.