A rectangle in the first quadrant has one vertex as the origin, and other vertex on the curve y= -x^2 -2x +4. If the rectangle has one side lying on the x-axis, find the largest possible area of the rectangle.

Answer 1

let the vertex which lies on the parabola be (x,y)
So the base of the rectangle is x
and its height is y
Area = xy
but y = -x^2 - 2x + 4
Area = x(-x^2 - 2x + 4)
= -x^3 - 2x^2 + 4x
d(Area)/dx = -3x^2 - 4x + 4 = 0 for a max area
3x^2 + 4x - 4 = 0
(3x-2)(x+2) = 0
x = 2/3 or x is a negative, which is no good

Max area = -(2/3)^3 - 2(2/3)^2 + 4(2/3)
= 40/27

Answer 2

square