Min(PA⋅PB) = ____.?

△APB = (3*7)/2 = 21/2 = (1/2)*PA⋅PB*sin(∠APB)
Max of sin(∠APB) = 1
Min of PA⋅PB = 21.

Assume that AB is on x-axis and (0,0) is the midpoint of AB .
So xy-coordinates of A,B,P are A(-7/2 , 0) , B(7/2 , 0) and P(x , 3) .

PA = √[(x + 7/2)^2 + 3^2]
= √[(x^2 + 7x + 49/4) + 9]
= √[(x^2 + 85/4) + 7x]

PB = √[(x - 7/2)^2 + 3^2]
= √[(x^2 - 7x + 49/4) + 9]
= √[(x^2 + 85/4) - 7x]

PA*PB = √[[(x^2 + 85/4) + 7x][(x^2 + 85/4) - 7x]]
= √[(x^2 + 85/4)^2 - 49x^2]
= √[x^4 + (85/2)x^2 + 7225/16 - 49x^2]
= √[x^4 - (13/2)x^2 + 7225/16]
= √[(x^2 - 13/4)^2 - 169/16 + 7225/16]
= √[(x^2 - 13/4)^2 + 441]

Therefore , Min(PA*PB) = √441 = 21 .
(At x^2 = 13/4 , x = ±1.8027... )

85/4