Results 1 to 7 of 7
Like Tree1Thanks
  • 1 Post By Reckoner

Thread: Pentagon Problem

  1. #1
    Newbie
    Joined
    Jun 2012
    From
    Georgia
    Posts
    24

    Pentagon Problem

    Pentagon Problem-tr.jpg

    Pentagon ABCDE

    BC = CD = BD = AE

    Perimeter (ABCDE) = 10

    Find the sides of rectangle, for which the area of the pentagon will be maximum.

    Please help! i need it so much

    (sry for my english)
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Jun 2012
    From
    AZ
    Posts
    616
    Thanks
    97

    Re: Pentagon Problem

    Let $\displaystyle BC = x$ and $\displaystyle AB = y$. The area of triangle BCD is $\displaystyle \frac{x^2 \sqrt{3}}{4}$, and the area of ABDE is $\displaystyle xy$. Hence the area of the pentagon P is


    $\displaystyle [P] = \frac{x^2 \sqrt{3}}{4} + xy = \frac{x^2 \sqrt{3} + 4xy}{4}$


    However you know that the perimeter is 10, so $\displaystyle 3x + 2y = 10 \Rightarrow y = \frac{10-3x}{2}$. Substitute into the area equation to obtain


    $\displaystyle [P] = \frac{x^2 \sqrt{3} + 4x(\frac{10-3x}{2}))}{4}$


    Simplify, and differentiate both sides with respect to x and find critical points.
    Last edited by richard1234; Jun 19th 2012 at 12:24 PM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Jun 2012
    From
    Georgia
    Posts
    24

    Re: Pentagon Problem

    the answer is 10/(6-sqr(3)) and (15-5sqr(3))/(6-sqr(3)) but i cant get those answers
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor Reckoner's Avatar
    Joined
    May 2008
    From
    Baltimore, MD (USA)
    Posts
    1,024
    Thanks
    76
    Awards
    1

    Re: Pentagon Problem

    Quote Originally Posted by Telo View Post
    Find the sides of rectangle, for which the area of the pentagon will be maximum.

    Please help! i need it so much
    Suppose that the vertical sides of the rectangle each have length $\displaystyle a,$ and suppose that the other sides have length $\displaystyle b.$

    The area of the pentagon is

    $\displaystyle A = ab + \frac12b\left(\frac{\sqrt3}2b \right)$

    $\displaystyle \Rightarrow A = ab + \frac{b^2\sqrt3}4,$

    and the perimeter is

    $\displaystyle P = 2a + 3b = 10$

    $\displaystyle \Rightarrow b = \frac13\left(10 - 2a\right).$

    Substituting this into the area equation above produces

    $\displaystyle A = a\left[\frac13(10-2a)\right] + \left[\frac13(10-2a)\right]^2\frac{\sqrt3}4$

    $\displaystyle = \frac19\left[\left(\sqrt3-6\right)a^2 + \left(30 - 10\sqrt3\right)a + 25\sqrt3\right]$

    Differentiating,

    $\displaystyle \frac{dA}{da} = \frac19\left[\left(2\sqrt3 - 12\right)a + 30 - 10\sqrt3\right]$

    We locate the critical value:

    $\displaystyle \frac{dA}{da} = 0$

    $\displaystyle \Rightarrow\frac19\left[\left(2\sqrt3 - 12\right)a + 30 - 10\sqrt3\right]=0$

    $\displaystyle \Rightarrow\left(2\sqrt3 - 12\right)a = 10\sqrt3 - 30$

    $\displaystyle \Rightarrow a = \frac{10\sqrt3 - 30}{2\sqrt3 - 12} = \frac{5\sqrt3 - 15}{\sqrt3 - 6}$

    $\displaystyle \Rightarrow a = \frac{15 - 5\sqrt3}{6 - \sqrt3}$

    Now you can find $\displaystyle b.$
    Last edited by Reckoner; Jun 19th 2012 at 11:25 AM.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor Reckoner's Avatar
    Joined
    May 2008
    From
    Baltimore, MD (USA)
    Posts
    1,024
    Thanks
    76
    Awards
    1

    Re: Pentagon Problem

    Quote Originally Posted by richard1234 View Post
    However you know that the perimeter is 10, so $\displaystyle 4x + 2y = 10 \Rightarrow y = 5-2x$.
    Small correction: that should be $\displaystyle 3x + 2y = 10.$ We want the perimeter of the pentagon, so we just need the outer sides.
    Thanks from richard1234
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Newbie
    Joined
    Jun 2012
    From
    Georgia
    Posts
    24

    Re: Pentagon Problem

    thanks alot!
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Super Member
    Joined
    Jun 2012
    From
    AZ
    Posts
    616
    Thanks
    97

    Re: Pentagon Problem

    @Reckoner whoops. Can't do math in my head anymore lol. I just fixed my original post.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. pentagon
    Posted in the Geometry Forum
    Replies: 5
    Last Post: Mar 11th 2011, 04:15 PM
  2. Pentagon problem
    Posted in the Algebra Forum
    Replies: 4
    Last Post: May 10th 2010, 08:49 AM
  3. Pentagon
    Posted in the Math Challenge Problems Forum
    Replies: 3
    Last Post: Jun 29th 2009, 08:09 PM
  4. Pentagon Problem
    Posted in the Geometry Forum
    Replies: 1
    Last Post: May 1st 2009, 11:40 AM
  5. Pentagon
    Posted in the Geometry Forum
    Replies: 1
    Last Post: Aug 28th 2008, 11:10 PM

Search Tags


/mathhelpforum @mathhelpforum