ABC is an isosceles right angle triangle with AB=BC =1.If P is a point on AC then min(max(areaAPQ,(areaPQBR),area PRC))

i am getting answer 2/9. i have done using graph, take one variable x and then express other lengths in terms of that. make three graphs of x(1-x), x^2/2 and (1-x)^2/2, using which plot the function g(x)= max[x(1-x), x^2/2, (1-x)^2/2]. its minimum value will be blatantly obvious from the graph, occur at x=1/3

and thence min(max(areaAPQ,(areaPQBR),area PRC))= 2/9