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

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

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 $\displaystyle k > 0$

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

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

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

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

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

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

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

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

3. Exactly what I needed! Thanks.