Frank Chance     Team 3  

A piece of wire of 10 m is bent into the shape of a square and a 
circle.  Which combination will yield a) maximum area b)minimum area?

a)  Area = L^2 + (Pi)*r^2		Perimeter = 4*L + 2*(Pi)*r
						10 = 4L + 2*(Pi)*r
						(10-2*(Pi)*r)/4 = L, so
						L = (5/2) - (Pi*r)/2
A = (2.5-(Pi*r)/2)*(2.5-(Pi*r)/2)+(Pi*r^2)
A = 6.25 - 2.5*Pi*r + .25*Pi^2*r^2 + Pi*r^2
so A' = 0 = (-2.5)*Pi + .5*Pi^2*r + 2*Pi*r     

A'' = (+), so this is a minimum

solve for r, which is( r = 5/(Pi + 4)), and substitute into 
Perimeter equation.
solve for L, then, which is (L = 10/(Pi + 4)).  So, one would cut the 
wire @ (4*L) from the end.

b)
	Look at the graph of the area function, with r substituted in.  
The graph is a parabola and increases to r=infinity.  Therefore, the 
largest possible area is that where the whole wire is bent into a 
circle.