# Thread: Optimizing area within a parabola

1. ## Optimizing area within a parabola

Find the largest rectangle that is inside the graph of the parabola $y = x^2$ below the line $y = a$ (a is an unspecified constant value), with the top side of the rectangle on the horizontal line $y = a$.
The explanation says we want $x$ to be in $[0, \sqrt{a}]$. What does $\sqrt{a}$ mean?

2. ## Re: Optimizing area within a parabola

Is it the point where $y=a$ intersects the parabola?

We set $x^2 = a$ and then solve for $x$?

3. ## Re: Optimizing area within a parabola

Originally Posted by maxpancho
The explanation says we want $x$ to be in $[0, \sqrt{a}]$. What does $\sqrt{a}$ mean?
??? The square root of a, of course.

Is it the point where y=a intersects the parabola?

We set x 2 =a and then solve for x ?
Yes, the line y= a intersects $y= x^2$ where $x= \sqrt{a}$. Since the given region is to lie inside (above) the graph of $y= x^2$ and below $y= a$, any point in the region must have $-\sqrt{a}\le x\le \sqrt{a}$.

They are probably assuming a few things not explicitly stated: The largest rectangle will have sides parallel to the x and y axes. The top edge will be on the line y= a. The largest rectangle will be symmetric about y= 0. With a little thought you should see that those things follow from "symmetry conditions". For example, if a rectangle has vertical sides from -a to b with b< a, the we can get a larger rectangle by extending the right side from b to a.

Given that, such a rectangle must have vertices at (-x, a), (x, a), on the line y= a, and, going straight down to the parabola, $(-x, x^2)$, $(x, x^2)$. The horizontal sides of that rectangle, from x to -x, have length 2x and the vertical sides. from a to $x^2$, have length [tex]a- x^2[tex] The area to be maximized is $(2x)(a- x^2)= 2ax- 2x^3$.