Results 1 to 5 of 5
Like Tree2Thanks
  • 1 Post By Prove It
  • 1 Post By Prove It

Math Help - Heron's Forumla for right angled triangles

  1. #1
    Member
    Joined
    Mar 2009
    Posts
    182
    Thanks
    1

    Heron's Forumla for right angled triangles

    Hi everyone, im trying to do a question involving Heron's Formula where,

    A=\sqrt {s \left( s-a \right)  \left( s-b \right)  \left( s-c \right)}

    and s = (a + b + c)/2

    Using Lagrange Multipliers, i'm trying to find in terms of the fixed perimeter P = 2s, the maximum area of aright-angled triangle with perimeter P.

    There's a pretty standard proof out there for Heron's formula to maximise an equilateral triangle, so im trying to adapt it for a right angled triangle by labelling the sides "h", "hsin(t)" and "hcos(t)" where one of the interior angles of the triangle is "t". Is this even the right way to go? I get pretty stuck at this point anyway, trying to produce gradients, and the consequencial Lagrange Multipliers.

    Anyone able to get me stated on this?

    Thanks
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,560
    Thanks
    1425

    Re: Heron's Forumla for right angled triangles

    Well for one thing, since it's a right angle triangle, you know that $\displaystyle \begin{align*} c = \sqrt{a^2 + b^2} \end{align*}$, giving $\displaystyle \begin{align*} P = a + b + \sqrt{a^2 + b^2} \end{align*}$. Surely this must be easier to work with as it is now only in two variables...
    Thanks from sirellwood
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Mar 2009
    Posts
    182
    Thanks
    1

    Re: Heron's Forumla for right angled triangles

    Ok that makes sense, so now we are only working with hsin(t) and hcos(t)....

    My confusion still remains though of how to proceed from here using Lagrange multipliers. Don't I have to create a function of a and b and work out the gradients and then introduce lambda from there?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,560
    Thanks
    1425

    Re: Heron's Forumla for right angled triangles

    In fact, this seems to be quite an easy problem to set up.

    You want to maximise $\displaystyle \begin{align*} A(a, b) = \frac{1}{2}a\,b \end{align*}$ subject to $\displaystyle \begin{align*} P(a, b) = a + b + \sqrt{a^2 + b^2} = P \end{align*}$, where P is some fixed value.

    So now we want to solve the system

    $\displaystyle \begin{align*} \nabla A(a, b) &= \lambda \nabla P(a, b) \\ P(a, b) &= P \end{align*}$

    It should be clear that $\displaystyle \begin{align*} \lambda \neq 0 \end{align*}$ as it would end up with 0 area, clearly not a maximum.

    So expanding the partial derivatives we get

    $\displaystyle \begin{align*} \frac{\partial A(a, b)}{\partial a} &= \lambda \, \frac{\partial P(a, b)}{\partial a} \\ \frac{\partial A(a, b)}{\partial b} &= \lambda \, \frac{\partial P(a, b)}{\partial b} \\ P(a, b) &= P \\ \\ \\ \frac{1}{2}b &= \lambda \left( 1 + \frac{a}{\sqrt{a^2 + b^2}} \right) \\ \frac{1}{2}a &= \lambda \left( 1 + \frac{b}{\sqrt{a^2 + b^2}} \right) \\ a +b + \sqrt{a^2 + b^2} &= P \\ \\ \\ \frac{\frac{1}{2}b}{\frac{1}{2}a} &= \frac{\lambda \left( 1 + \frac{a}{\sqrt{a^2 + b^2}} \right) }{\lambda \left( 1 + \frac{b}{\sqrt{a^2 + b^2}} \right) } \\ \\ \frac{b}{a} &= \frac{a + \sqrt{a^2 + b^2}}{b + \sqrt{a^2 + b^2}} \\ \\ b \left( b + \sqrt{a^2 + b^2} \right) &= a \left( a + \sqrt{a^2 + b^2} \right) \\ \\ b^2 + b\sqrt{a^2 + b^2} &= a^2 + a\sqrt{a^2 + b^2} \\ \\ 0 &= a^2 - b^2 + a\sqrt{a^2 + b^2} - b\sqrt{a^2 + b^2} \\ \\ 0 &= \left( a - b \right) \left( a + b \right) + \left( a - b \right) \sqrt{a^2 + b^2} \\ \\ 0 &= \left( a - b \right) \left( a + b + \sqrt{a^2 + b^2} \right)\end{align*}$

    So $\displaystyle \begin{align*} a - b = 0 \implies a = b \end{align*}$ or $\displaystyle \begin{align*} a + b + \sqrt{a^2 + b^2} = 0 \end{align*}$ for a maximum. But $\displaystyle \begin{align*} a + b + \sqrt{a^2 + b^2} = P \end{align*}$, which we obviously don't want to be 0 for a maximum.

    So the maximum area must occur when $\displaystyle \begin{align*} a = b \end{align*}$, giving

    $\displaystyle \begin{align*} a + b + \sqrt{a^2 + b^2} &= P \\ a + a + \sqrt{a^2 + a^2} &= P \\ 2a + \sqrt{2a^2} &= P \\ 2a + \sqrt{2}\,a &= P \\ a \left( 2 + \sqrt{2} \right) &= P \\ a &= \frac{P}{2 + \sqrt{2}} \\ a &= \frac{ \left( 2 - \sqrt{2} \right) P }{2} \end{align*}$

    and thus the maximum area is

    $\displaystyle \begin{align*} A &= \frac{1}{2}a\,b \\ &= \frac{1}{2}a^2 \\ &= \frac{1}{2} \left[ \frac{ \left( 2 - \sqrt{2} \right) P }{2} \right] ^2 \\ &= \frac{1}{2} \left[ \frac{ \left( 2 - \sqrt{2} \right) ^2 \, P^2}{4} \right] \\ &= \frac{\left( 2 - \sqrt{2} \right) ^2 \, P^2}{8} \\ &= \frac{ \left( 4 - 4\sqrt{2} + 2 \right) P^2 }{8} \\ &= \frac{\left( 3 - 2\sqrt{2} \right) P^2 }{4} \end{align*}$
    Thanks from sirellwood
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member
    Joined
    Mar 2010
    Posts
    980
    Thanks
    236

    Re: Heron's Forumla for right angled triangles

    Impressive!

    - Hollywood
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. NON right-angled triangles - Quick pic
    Posted in the Trigonometry Forum
    Replies: 3
    Last Post: January 16th 2012, 12:17 PM
  2. Heron's forumla
    Posted in the Geometry Forum
    Replies: 1
    Last Post: November 2nd 2010, 08:37 PM
  3. Trig / right angled triangles / SIN COS TAN
    Posted in the Trigonometry Forum
    Replies: 11
    Last Post: May 8th 2010, 04:20 AM
  4. Right-Angled Triangles...
    Posted in the Trigonometry Forum
    Replies: 3
    Last Post: May 3rd 2010, 06:31 AM
  5. Replies: 8
    Last Post: March 31st 2009, 05:53 AM

Search Tags


/mathhelpforum @mathhelpforum