Question 606207
This has MANY answers....
We know that the area of a rectangle is 
{{{A=L*W}}}   
So we have:
{{{59=L*W}}}
{{{L=59/W}}}
But we can substitute in any number for L and get a corresponding W.  (If it was a square however, there would only be one answer as we substitute L=W and solve for W.  

With the perimeter the same problem remains.  Once we have a L and W we have:
{{{P=2L+2W}}}
But due to the fact that there are many such L and W, there are just as many perimeters.   


Hopefully this clears things up a bit, if there was any more specifications with the question Let me know so I can be more clear with my answer.  



Okay, using the new information, if we want the smallest rectangle with whole number lengths and widths-- we know that the smallest length we can have is 1 making the width 59 (not that this is smaller perimeter than if we had a square with area 59 and any other rectangles with that area with whole number lengths).   

From here, you can use the above formula and find perimeter.  :)