Think of the set of rectangles with integer sides such that each rectangle
has its length and width together made up of 10 segments of sizes
1,2,3,...,10. So its perimeter = 2L+2W 2(L+W)= 2(1+...+10) = 110.
So our problem is a finite subset of the set of areas of the infinite set of
rectangles with perimeter 110.  We know that the area of the largest
possible rectangle is a square with sides 1/4 of 110 or 27.5.  So to find
the maximum area in our set of areas we need to find a rectangle which is as
close as possible to a square with sides 27.5. 

So I'll see if it's possible to get one dimension to be 27 and the other to
be 28, then they'll be closest together.  I'll start with the largest
possible two (10+9) and the smallest possible two (1+2) and get
(10+9+1+2)=22, so if I add 5 to that I'll have 10+9+5+1+2=27.  Then the
other five will be 8+7+6+4+3=28.  Hurray! So I claim that the largest
product is 27*28 = 756.  For the two can be no closer in area to the area of
the square with maximum area, which is 27.5×27.5.
  
So the maximum product possible is (10+9+5+1+2)(8+7+6+4+3) = (27)(28) = 756.

The minimum area of a rectangle with circumference 110 when one
of the two dimensions is 0 and the other is 55, which is a rectangle
degenerated into a line segment.  That is when the two dimensions are as far
apart as possible.

So the minimum product is when one factor is as small as possible and the
other is as large as possible.  That's this case:

(1+2+3+4+5)(6+7+8+9+10) = (15)(40) = 600

So 600 is the minimum product

Answers: the maximum value of P = 756 (when the two numbers are closest
together, (27)(28) = 756
         the minimum value of P = 600 (when the two numbers are farthest
apart, (15)(40) = 600 

Edwin