A rectangle is inscribed in parabola y^2=8x with one of its vertical side passes through the point (6,0). Find its maximum area

Answer 1

y^2 = 8x is a rightward opening parabola with vertex at the origin. At x=6, y = sqr48 = + or - 4sqr3

a rectangle inside the parabola would have two vertices of (6,4sqr3) and (6,-4sqr3)

It's area would be 8sqr3 times it's length which could be infinite, with no maximum other than infinity.

If it's a square it's area would be 8sqr3 squared = 64x3 = 192

Or if maybe you meant the parabola is inscribed inside the rectangle, the rectangle would be 6 by 8sqr3 = 48sqr3, with left side bounded by x=0, right side x=6, top y=4sqr3, bottom y=-4sqr3