Results 1 to 3 of 3

Thread: Optimization Problem

  1. #1
    Member
    Joined
    Mar 2010
    Posts
    78

    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)?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    16,216
    Thanks
    3702
    Quote Originally Posted by RogueDemon View Post
    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 = $\displaystyle \dfrac{k^3 \sqrt{3}}{18}$
    Attached Thumbnails Attached Thumbnails Optimization Problem-parabolarectangle.jpg  
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Mar 2010
    Posts
    78
    Exactly what I needed! Thanks.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Optimization Problem.
    Posted in the Calculus Forum
    Replies: 1
    Last Post: May 5th 2010, 01:44 AM
  2. Another Optimization problem
    Posted in the Calculus Forum
    Replies: 5
    Last Post: Oct 26th 2008, 08:58 PM
  3. Help with optimization problem
    Posted in the Calculus Forum
    Replies: 4
    Last Post: Nov 6th 2007, 07:20 PM
  4. Optimization Problem
    Posted in the Calculus Forum
    Replies: 5
    Last Post: Apr 8th 2007, 08:30 PM
  5. Optimization Problem.
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Apr 8th 2007, 06:38 PM

Search tags for this page

Click on a term to search for related topics.

Search Tags


/mathhelpforum @mathhelpforum