Thread: Max/Min Problem...Having trouble with it?

1. Max/Min Problem...Having trouble with it?

I have absolutely no idea how to approach this question...

Draw a diagram showing the region enclosed between the parabola y^2=4ax and its latus rectum x=a

The, find the dimensions of the rectangle of max area that can be inscribed in this region.

I drew the diagram...but so far all i have for the area equation is A=2yx (y is doubled because the parabola extends below the axis as well XD)

Im having trouble uploading a picture, so ill describe the diagram...its a sideways parabola extending on the right side of the y axis, and the latus rectum cuts the x axis at "a" and the parabola when y= 2a and -2a

Basically, i have no idea what to do...could anyone help?

2. The length along the x-axis is going to be $\displaystyle a-x$ for some x. The length along the y-axis is $\displaystyle 2\sqrt{4ax}=4\sqrt{ax}$.

That gives us the area $\displaystyle A(x)=4\sqrt{ax}(a-x)$ to maximize.