A rectangle is to be placed in the first quadrant, with one side on the x-axis and one side on the y-axis so that the rectangle lies below the parabola y=-3x^2+4. What is the maximum area of this rectangle?

a) 16/9

b) 3

c) 2/3

d) 2

Julius Mar 30, 2018

#1**+2 **

Mmmm...let's see if I know this one.....

Let the width of the rectangle = x

And let the height = -3x^2 + 4

So....the area, A = width * height = x (-3x^2 + 4) = -3x^3 + 4x

Take the derivative of A and set to 0

A' = -9x^2 + 4

-9x^2 + 4 = 0

-9x^2 = -4 divide both sides by -9

x^2 = 4/9 take the positive root

x = 2/3 = width

And the height is given by:

-3(2/3)^2 + 4 =

-3(4/9) + 4

-4/3 + 4

8/3

So....the max area is

(2/3)(8/3) = 16/9 units^2

CPhill Mar 30, 2018