+956 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
+129852 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
