Optimization: maximize a triangle surface

[Telemachus]

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 here is ok? Greetings.

 

[micromass]

Yes, what you're doing is good. And since f is an ugly function, I can imagine that the partial derivatives of f are very ugly to...

 

[Telemachus]

Thanks. I know that I must arrive to x=z=y, an isosceles triangle, but it seems to be harder than what I thought.

$$A^2=p(p^2-py-xp+xy)(p-z)=p^4-p^3y-p^3x+p^2xy-p^3z+p^2zy+p^2zx-pzxy$$
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.

 

[micromass]

At the end, you have to discriminate between to cases: x=y and x is not y. In the last case, you indeed have

$$z=\frac{2x^2-2y^2}{2x-2y}=x+y$$

Together with x+y+z=2p, this yields that z=p. And thus this gives the point (0,0,p).

 

[Telemachus]

I think I've arrived at this conclusion before following another way. But I thought I was wrong, because I've arrived to x=p, z=y=x, and then I thought that what I had was a point, not a surface, so I thought I had a minimum there, not a maximum. And here I think the same, if we got z=p then A=0 from here: $$A=\sqrt[ ]{p(p-x)(p-y)(p-z)}$$

I think the solution must be $$x=y=z\neq{p}$$

 

[Dick]

It will probably make things a lot clearer if you don't expand all of those products. It looks to me like your f_x equation is -(y-p)(z-p)=lambda/2 and your f_y equation is -(x-p)(z-p)=lambda/2. I think you can draw an important conclusion from those two alone.

 

[Telemachus]

Alright, thanks Dick, I'll try that way, and then I'll tell you :P

 

[D H]

I think you can draw an important conclusion from those two alone.

Exactly. Since the perimeter P is a given constant and x+y+z=P, there are only two degrees of freedom here.

 

[Telemachus]

I thought so, I think I did something like this in calculus 1 but with a square, that's what's in part drives me to think that the isosceles triangle is the one I'm looking for. But anyway this case seems complicated, maybe it is because I'm not seeing it clearly.

 

[Dick]

I thought so, I think I did something like this in calculus 1 but with a square, that's what's in part drives me to think that the isosceles triangle is the one I'm looking for. But anyway this case seems complicated, maybe it is because I'm not seeing it clearly.

It LOOKS complicated because you are multiplying everything out. Don't do that. It's not complicated.

 

[Telemachus]

This is what I have without the multiplications:

$$\begin{Bmatrix}-p(p-y)(p-z)=\displaystyle\frac{\lambda}{2}\\-p(p-x)(p-z)=\displaystyle\frac{\lambda}{2}\\-p(p-x)(p-y)=\displaystyle\frac{\lambda}{2}\\p=\displaystyle \frac{x+y+z}{2} \end{matrix}$$

Then I did: $$-p(p-y)(p-z)=\displaystyle\frac{\lambda}{2}\rightarrow{}z=\displaystyle\frac{\lambda}{2p(p-y)}+p$$

$$p=\displaystyle \frac{x+y+z}{2}\rightarrow{}z=-x-y+2p$$

$$-x-y+2p=\displaystyle\frac{\lambda}{2p(p-y)}+p\rightarrow{}\displaystyle\frac{\lambda}{2}=(-x-y+p)p(p-y)$$

Then:
$$(-x-y+p)p(p-y)=-p(p-x)(p-y)\rightarrow{}-p+x=-x-y+p\rightarrow{}2x=-y+2p\rightarrow{}x=\displaystyle\frac{-y}{2}+p$$

$$z=-x-y+2p\rightarrow{}z=\displaystyle\frac{y}{2}-p-y+2p\rightarrow{}z=\displaystyle\frac{-y}{2}+p\therefore{z=x}$$

Now the problem is "y".
$$p=\displaystyle \frac{x+y+z}{2}\rightarrow{}y=2p-2x$$ it should give equal to x and z...

 

[Telemachus]

Ok, I did something really stupid there. I've found the way. Thank you all guys.

 

[Dick]

Ok, I did something really stupid there. I've found the way. Thank you all guys.

You're welcome and glad you got it. But I still don't think you are seeing the easy way. Look at the first two derivative equations. Don't they tell you really directly that (p-x)=(p-y)?

 

[Telemachus]

Haha yes, that's what I saw after I've posted here :P