# Heron's Forumla for right angled triangles

• May 2nd 2014, 05:36 AM
sirellwood
Heron's Forumla for right angled triangles
Hi everyone, im trying to do a question involving Heron's Formula where,

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

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
• May 2nd 2014, 06:32 AM
Prove It
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...
• May 2nd 2014, 06:54 AM
sirellwood
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?
• May 2nd 2014, 07:07 AM
Prove It
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*}
• May 2nd 2014, 07:52 PM
hollywood
Re: Heron's Forumla for right angled triangles
Impressive!

- Hollywood