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I thought I could use this $r$, being the circumradius of the triangle, thereby the area being: $\frac{abc}{4r}$. Now from here I need to find a,b,c so the area is maximal. Could I do this with Lagrange coefficients adding a requirement? I dont know which though.. This question was asked when doing conditional extremes..

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  • $\begingroup$ Are you asking only about the method of Lagrange coefficients, or would you also want a simpler method? $\endgroup$ Mar 21, 2016 at 21:44
  • $\begingroup$ Method of Lagrange coefficients and the other one for sport if possible. :) $\endgroup$
    – Jerry West
    Mar 21, 2016 at 22:23

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We define function $f$ which calculates triangle area by its angles like this: $(A\,,B\,,C \lt\pi)$ $$f(A,B,C)=2R^2\cdot\sin A\cdot\sin B\cdot \sin C$$ And function $g$ like this: $$g(A,B,C)=A+B+C=\pi$$ Now we need to solve: $$\cos A\cdot2R^2\sin B\cdot\sin C=\lambda\qquad (1)\\[2ex] \cos B\cdot2R^2\sin C\cdot\sin A=\lambda\qquad (2)\\[2ex] \cos C\cdot2R^2\sin A\cdot\sin B=\lambda\qquad (3)\\[2ex] A+B+C=\pi\qquad\qquad\qquad\quad (4)$$ Setting $(1)\,,(2)\,,(3)$ equal will show us: $$\cos A\cdot\sin B=\cos B\cdot\sin A\Rightarrow \sin(A-B)=0\\[2ex] \cos B\cdot\sin C =\cos C\cdot\sin B\Rightarrow \sin(B-C)=0\\[2ex] \cos C\cdot\sin A=\cos A\cdot\sin C\Rightarrow \sin(C-A)=0\\[2ex]$$ And since $A\,,B\,,C$ are angles of a triangle we can conclude: $$A=B=C=\frac{\pi}{3}\qquad,\lambda=\frac{3R^2}{4}$$ Therefore the triangle with maximal area is the equilateral triangle.

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