<PRE><B>Question 890284</B>
If all you are given is the perimeter then you can have several areas, one of which will be the maximum area and one of which will be the minimum area.
I am assuming that the length and width both have to be integers.
If not, then the number of possibilities will be endless.


For Example:


2L + 2W = P
L = length
W = width
P = perimeter = 36


formula becomes:
2L + 2W = 36
solve for L to get:
L = 18 - W


start from a width of 1 and work your way up to get the area and the perimeter each time based on the formula that L = 18 - W


you will get:


&lt;pre&gt;


W     L     P     A
1     17    36    17
2     16    36    32
3     15    36    45
4     14    36    56
5     13    36    65
6     12    36    72
7     11    36    77
8     10    36    80
9     9     36    81 ***** maximum area when L = W
10    8     36    80
etc...
they repeat from here on, i.e. 8*10 = 10*8, 7*11 = 11*7, etc.


&lt;/pre&gt;


Your maximum area is when L and W are closest in value to each other.
Your minimum area is when L and W are furthest apart in value from each other.


</PRE>