ABC is an isosceles right angled triangle with AB = BC = 1. If P is any point on BC, then min {max{(area APQ), (area PQBR), area PRC)}} is

The correct answer is:

2/9
Length of AC = √2
Let the length of AP be x
Length of PC= √2-x
Let angle QAP=angle BCP = z
Length of AQ = APcos 45 = AP/√2
Length of PQ= AP/ √2
Area of APQ = x²/4

Area of PQBR = PQ x PR = (x/√2 )(1-x/√2)
=x/√2-x²/2

Area of PRC = ½(PR)(RC)=1/2(1-x/√2)(1-x/√2)
= x²/4 -x/√2+1/2


From the area functions, we can see that max(Area APQ, area PQBR, area PRC)= area of APQ = x²/4 when x>=2√2/3
Minimum of the areas = area of PQBR =  x/√2- x²/2= 2/9

the minimum function gives us the minimum of the values passed to it . similarly, the max or maximum function returns the maximum of the values passed to it.