# Optimization: maximize a triangle surface

• Oct 31st 2010, 05:57 PM
Ulysses
Optimization: maximize a triangle surface
See if anyone can help me with this: Among all triangles of perimeter equal to P, find the one with the largest area. (Hint: use the formula $A=\sqrt[ ]{p(p-x)(p-y)(p-z)}$ where $P=2p$, $P$ is the perimeter).

So, I have $f|_s$, I think that must be solved using Lagrange multipliers, at least I don't see any other way.

I've proceeded this way: $f=\sqrt[ ]{p(p-x)(p-y)(p-z)}$, $s=p=\displaystyle\frac{x+y+z}{2}$

Well, I have done so, but all derivatives did wrong (I did $A^2$ arising as if $f=A^2$ and then apply the multiplier to with the 4 conditions ), it became ugly, maybe it was because of that. Anyway, would you tell me if what I did until here is ok? Greetings.
• Oct 31st 2010, 06:13 PM
HallsofIvy
First, since $\sqrt{x}$ is an increasing function, any maximum value of $\sqrt{x}$ is a maximum value of x so it is sufficient to maximize $A^2= p(p-x)(p- y)(p- z)$. Taking, as constraint, x+ y+ z= 2p, a constant, you want to maximize F(x, y, z)= p(p- x)(p- y)(p- z) with constraint G(x, y, z)= x+ y+ z= 2p. You want to find x, y, z such that $\nabla F= \lambda\nabla G$ for some constant $\lambda$. Don't forget that p is a constant.
• Oct 31st 2010, 06:57 PM
Ulysses
Lets see then if what I did its ok:

$A^2=p(p^2-py-xp+xy)(p-z)=p^4-p^3y-p^3x+p^2xy-p^3z+p^2zy+p^2zx-pzxy$
Then I call: $f=p(p^2-py-xp+xy)(p-z)=p^4-p^3y-p^3x+p^2xy-p^3z+p^2zy+p^2zx-pzxy$

And $s=p=\displaystyle\frac{P}{2}=\displaystyle\frac{x+ y+z}{2}$

$\begin{Bmatrix}f_x=p(py+pz-zy-p^2)\\f_y=p(px+pz-zx-p^2)\\f_z=p(px+py-p^2-xy)\end{matrix}$

$p\neq{0}$
Then:
$\begin{Bmatrix}py+pz-zy-p^2=\displaystyle\frac{\lambda}{2}\\px+pz-zx-p^2=\displaystyle\frac{\lambda}{2}\\px+py-p^2-xy=\displaystyle\frac{\lambda}{2}\\p=\displaystyle \frac{x+y+z}{2} \end{matrix}$

From the last condition I replaced p to get lambda.
Then I've found that:
$\displaystyle\frac{\lambda}{2}=-x^2+y^2+z^2-2zy$
Then I replace in one of the equations

$-x^2+y^2+z^2-2zy=\displaystyle\frac{(x+y+z)}{2}x+\displaystyle\ frac{(x+y+z)}{2}z-zx-(\displaystyle\frac{(x+y+z)}{2})^2$
So I get into what I think its an absurd:
$z=\displaystyle\frac{2y^2-2x^2}{2y-2x}$
I think its an absurd, but I haven't go ahead from hear. And the reason is that I think I should have an isosceles triangel, so I would be dividing be zero on this equality.