Question 985275
I'm assuming this rectangle has one edge on the x axis. Let P be one point of the rectangle. Also, let P be on the positive x axis to the left of the root of the given parabola.


The distance from (0,0) to P is some unknown quantity x. Due to symmetry, the edge of the rectangle that rests on the x axis is going to be 2*x units long.


The height of this rectangle will be -x^2+12 because the upper corner point is on the parabola (draw a picture and this should be clear hopefully).


So we have length = 2x and width = -x^2 + 12


The area of this rectangle is...


A = L*W
A = (2x)*(-x^2 + 12)
A = -2x^3 + 24x


There are two ways to find the max area


1) Use a graphing calculator to find the peak max point on y = -2x^3 + 24x. This is like the vertex of a parabola, but not entirely. I like to think of it like it though. If you have a TI calculator, you can use the min/max feature to find the peak point.


2) Use calculus to find the max without using a graphing calculator. This is only if you're in a calculus class.


Using either method (I recommend method #1 since it sounds like you're in a high school algebra class), you should find that the max occurs at the point (2,32). 


So if x = 2, the length is 2*x = 2*2 = 4 and the width is -x^2 + 12 = -(2)^2 + 12 = 8. Notice how L*W = 4*8 = 32


The max area possible is 32 square units (length = 4, width = 8)