Let the height of the triangle be #h#.
Then half of the third side, #h#, and the given side (3) must form a right angle side. So. The base is given by #2times sqrt(3^2-h^2)#
So, the area of the isosceles triangle is #A = 1/2times h times (2times sqrt(3^2-h^2)) = h sqrt(9-h^2)#
So. #A# is the geometric mean of two positive quantities : #h^2# and #9-h^2#, which have a given arithmetic mean (namely 4.5). Since the geometric mean of two positive quantities is never larger than their arithmetic mean, the largest possible value for the area is 4.5.
The maximum area is achieved when #h=sqrt 4.5#
Alternate :
The area is given by #1/2 times 3 times 3 times sintheta# where #theta# is the vertex angle of the isosceles triangle. This obviously has the maximum value of 4.5 (when #theta = 90^circ#)
