1. ## Optimization Problem

6. The base of a rectangle lies along the x axis, and the upper two vertices are on the curve defined by y = kx - x^2. Determine the dimensions of the rectangle with the maximum area.

I started off with setting up the formula $A = 2lw = 2xy$ and the constraint $y = kx - x^2$. After subbing $y = kx - x^2$ into $A$, I found the derivative of $A$ and solved for $x$, which turned out to be $2/3k$. I then subbed $x$ into $y = kx - x^2$, which gave me a $y$ value of $2k^2/9$, and subsequently I found the maximum area to be $8k^3/27$.

The problem is that after I double checked my diagram, I found that it was inconsistent with the actual equation. Namely, it did not account for the fact that when x = 0, y = 0, and thus it had a large gap in the center.

So my question is this: Did I set up my equations correctly? Is my parabola supposed to look something like http://jwilson.coe.uga.edu/emt668/EM...s/image007.jpg or something like http://xsquared.wikispaces.com/file/...l_parabola.jpg (but moved to the right a bit more)?

2. Originally Posted by RogueDemon
6. The base of a rectangle lies along the x axis, and the upper two vertices are on the curve defined by y = kx - x^2. Determine the dimensions of the rectangle with the maximum area.

I started off with setting up the formula $A = 2lw = 2xy$ and the constraint $y = kx - x^2$. After subbing $y = kx - x^2$ into $A$, I found the derivative of $A$ and solved for $x$, which turned out to be $2/3k$. I then subbed $x$ into $y = kx - x^2$, which gave me a $y$ value of $2k^2/9$, and subsequently I found the maximum area to be $8k^3/27$.

The problem is that after I double checked my diagram, I found that it was inconsistent with the actual equation. Namely, it did not account for the fact that when x = 0, y = 0, and thus it had a large gap in the center.

So my question is this: Did I set up my equations correctly? Is my parabola supposed to look something like http://jwilson.coe.uga.edu/emt668/EM...s/image007.jpg or something like http://xsquared.wikispaces.com/file/...l_parabola.jpg (but moved to the right a bit more)?
w/o loss of generality, let $k > 0$

$y = kx - x^2$ is a parabola that opens downward w/ x-intercepts at $x = 0$ and $x = k$

base of inscribed rectangle = $k - 2x$ , $0 < x < \dfrac{k}{2}$

height of inscribed rectangle = $kx - x^2$

$A = (k-2x)(kx-x^2)$

$A = 2x^3 - 3kx^2 + k^2 x$

$\dfrac{dA}{dx} = 6x^2 - 6kx + k^2 = 0$

$x = \dfrac{k(3 - \sqrt{3})}{6}$

max area = $\dfrac{k^3 \sqrt{3}}{18}$

3. Exactly what I needed! Thanks.