A rectangle is inscribed in the interior section cut from the parabola x = 1/4 y2 by the line x=3. One side of the rectangle lies on the line. Fine the maximum area of such a rectangle.

Answer 1

Make a sketch.
Let the points on the parabola be P(x,y) and Q(x,-y)

Area = 2y(3-x)
= 2y(3-(1/4)y^2)
= 6y - (1/2)y^2
d(Area)/dy = 6 - y = 0 for a max of Area
y = 6

the max area is 6(6) - (1/2)(36)
= 18 square units

check my arithmetic