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The title is the text of the exercise (A is area, P is perimeter). I would like to solve it with Lagrange moltiplicators. I already found a solution with Erone's formula. But I would like to approach the problem describing the triangles with sine and cosine, this is my idea: I consider two circumferences with differents radiuses, circumference 1 has radius $\rho_1$ and the circumference 2 $\rho_2$, the radiuses changes and $0<\rho_1, \rho_2<1$. Now, these two circumferences have the same sine, so I'm describing all the possible triangles (the circumferences are connected by them sines) then $\rho_1\sin\theta=\rho_2\sin\phi$. Now it's easy to write the heights of the triangles because are equal to the sine (one of them), and for the perimeter (which is the bound) I can write: $$P=\rho_1\cos\theta+\rho_2\cos\phi+\rho_1+\rho_2=1$$ Also the area is, for example: $$A=(\rho_1\cos\theta+\rho_2\cos\phi)(\rho_1\sin\theta)$$

Now I can write Lagrange's equation for Lagrange moltiplicators: $$\mathscr L=A-\lambda P=(\rho_1\cos\theta+\rho_2\cos\phi)(\rho_1\sin\theta)-\lambda(\rho_1\cos\theta+\rho_2\cos\phi+\rho_1+\rho_2-1)$$

But I'm not considering that $0<\rho_1, \rho_2<1$ and it cames out, when I establish $\nabla\mathscr L=0$, that $\rho_1<0$

Is this approach totally didastrous or am I only making mistakes in resolution?

Thanks a lot

pter26
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  • What is the question? If $A=1$ then that is the max of $A$. Did you mean something else? – lulu Jul 22 '17 at 12:30
  • I'm sorry, I mean that perimeter is one, I'm going to edit the title – pter26 Jul 22 '17 at 12:31
  • That makes sense. here is a related question. At least one of the answers gives a non-Heron proof. – lulu Jul 22 '17 at 12:32
  • The triangle with max area is equilateral. Since the perimeter is $1$ the side is $\dfrac{1}{3}$ and the area is $A=\dfrac{\sqrt{3}}{36}$ – Raffaele Jul 22 '17 at 19:58

1 Answers1

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From the sine theorem: $$abc=8R^3\sin{\alpha}\sin{\beta}\sin{\gamma}, P=2R(\sin{\alpha}+\sin{\beta}+\sin{\gamma}).$$ The area is: $$A=\frac{abc}{4R}=\frac{P^2\sin{\alpha}\sin{\beta}\sin{\gamma}}{2(\sin{\alpha}+\sin{\beta}+\sin{\gamma})^2}\le \frac{P^2(\sin{\alpha}+\sin{\beta}+\sin{\gamma})}{2\cdot 27}\le \frac{P^2}{12\sqrt{3}}.$$ The equality occurs when $\alpha=\beta=\gamma=\frac{\pi}{3}$.

farruhota
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