SOLUTION: The perimeter of a rectangle is 2012, and lengths of all sides are integers. What is the smallest possible area of this rectangle?

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Question 570786: The perimeter of a rectangle is 2012, and lengths of all sides are integers. What is the smallest possible area of this rectangle?
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
If x and y are the dimensions of this rectangle, we know that
2%28x%2By%29=2012 so x%2By=2012%2F2 x%2By=1006 and y=1006-x
The area, as a function of x is
A%28x%29=%281006-x%29x
A%28x%29=-x%5E2%2B1006x
That quadratic equation represents a parabola.
Its axis of symmetry is x=-1006%2F%282%2A%28-1%29%29 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.