Results 1 to 3 of 3

Math Help - [SOLVED] Maximization of pentagon area...

  1. #1
    Member
    Joined
    Sep 2009
    Posts
    82

    [SOLVED] Maximization of pentagon area...

    A pentagon is formed by placing an isosceles triangle on a rectange. If the pentagon has fixed perimeter P, find the lengths of the sides of the pentagon that maximize the area of the pentagon.

    Check number 4 for a picture and answers: http://ocw.nctu.edu.tw/upload/calcul.../ca2_test5.pdf

    Its how to get the answers that i need help with...

    Maximize the area:
    A=xy+xtan(\theta)
    Constraint:
    P=x+2y+x^2+y^2

    using lagrange multipliers:
    <y+tan(\theta),x>=\lambda<1+2x,2+2y>

    y+tan(\theta)=\lambda(1+2x)
    x=\lambda(2+2y)

    From here i cannot solve for any one variable. Caused mainly by the tan(\theta), which adds one too many variables. I am most likely missing some little step to get me there, or mabye lagrange multipliers are not the best method to use...Please let me know what to do. Thanks!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member
    Joined
    Aug 2009
    Posts
    80
    Quote Originally Posted by snaes View Post
    A pentagon is formed by placing an isosceles triangle on a rectange. If the pentagon has fixed perimeter P, find the lengths of the sides of the pentagon that maximize the area of the pentagon.

    Check number 4 for a picture and answers: http://ocw.nctu.edu.tw/upload/calcul.../ca2_test5.pdf

    Its how to get the answers that i need help with...

    Maximize the area:
    A=xy+xtan(\theta)
    Constraint:
    P=x+2y+x^2+y^2

    using lagrange multipliers:
    <y+tan(\theta),x>=\lambda<1+2x,2+2y>

    y+tan(\theta)=\lambda(1+2x)
    x=\lambda(2+2y)

    From here i cannot solve for any one variable. Caused mainly by the tan(\theta), which adds one too many variables. I am most likely missing some little step to get me there, or mabye lagrange multipliers are not the best method to use...Please let me know what to do. Thanks!
    Assuming x is the width of the rectangle, y is the height of the rectangle and h is the height of the isosceles triangle, then A = xy + xh/2, P = x + 2y + 2sqrt((x/2)^2 + h^2) = x + 2y + sqrt(x^2 + 4h^2). Thus grad(A) = <y + h/2, x, x/2> and grad(P) = <1 + x/sqrt(x^2 + 4h^2), 2, 4h/sqrt(x^2 +4h^2)>. Setting grad(A) = lambda * grad(P), then this, together with the constraint on the perimeter, gives a system of four equations in four variables.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Sep 2009
    Posts
    82
    Ok, it was the tangent \theta that messed me up thanks. Now I got it replaced by h and it works out.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] Area maximization
    Posted in the Calculus Forum
    Replies: 4
    Last Post: October 15th 2011, 11:51 PM
  2. Problems involving maximization of Area
    Posted in the Geometry Forum
    Replies: 11
    Last Post: May 24th 2011, 12:31 PM
  3. Area of the pentagon
    Posted in the Geometry Forum
    Replies: 1
    Last Post: April 8th 2010, 05:37 PM
  4. Area of a Regular Pentagon
    Posted in the Trigonometry Forum
    Replies: 7
    Last Post: February 28th 2010, 12:27 PM
  5. Replies: 3
    Last Post: September 5th 2009, 05:21 PM

Search Tags


/mathhelpforum @mathhelpforum