Question 570786
If {{{x}}} and {{{y}}} are the dimensions of this rectangle, we know that
{{{2(x+y)=2012}}} so {{{x+y=2012/2}}} {{{x+y=1006}}} and {{{y=1006-x}}}
The area, as a function of x is
{{{A(x)=(1006-x)x}}}
{{{A(x)=-x^2+1006x}}}
That quadratic equation represents a parabola.
Its axis of symmetry is {{{x=-1006/(2*(-1))}}} {{{x=503}}}
The maximum area occurs at {{{x=503}}}, when the rectangle is a square.
Moving away from that point, to either side of {{{x=503}}}, the area decreases.
Since the length of the sides are integers, the minimum will be for {{{x=1}}} and {{{x=1005}}}, when one side measures 1 and the other 1005. It is the same solutionm no matter what side length we call x.
The minimum area is 1005.