show that out of all triangles inscribed in a circle the one with maximum area is equilateral show that out of all triangles inscribed in a circle the one with maximum area is equilateral
How do i start. I have to use function of two variables
Thanks
 A: I suggest you to consider a triangle, and show that if it isn't equilateral, you can find another triangle with a greater area.
That way you are sure that the equilateral triangle is the biggest you can get.
Hint:  

Consider the bisectors of the sides of your triangle.

Hint 2:

 If you move a vertex parallely to the opposite side, you dont change the area. So, if you move a vertex that isn't on the bisector to the bisector, you have the same area but the vertex you moved is now strictly inside the circle.

A: Most geometric treatments of this problem tacitly assume that a triangle of largest area in fact exists, and argue as follows: Since non-equilateral triangles can be transformed into inscribed triangles with larger area, the maximal triangle has to be equilateral. It is a nontrivial geometric exercice to provide a proof that any non-equilateral triangle has actually smaller area than the equilateral one.
A non-equilateral triangle has an angle $\alpha<60^\circ$. After a preliminary step we may assume that this triangle ABC is isosceles with base $BC$, see the following figure. In the same figure we draw the equilateral triangle  $AB'C'$.
From $\alpha<60^\circ$it follows that the point $B'$ lies strictly between the point $B$ and the point $S$ on the circle. This allows to draw the following conclusion:
$${\rm area\,}(ABC)<{\rm area\,}(AB'C)<{\rm area\,}(AB'C')\ ,$$
because at each step we make a triangle more isosceles, keeping two of ists vertices fixed.

A: The straightforward way to demonstrate this is the geometric approach, outlined in three different ways by the other solutions thus far.
If you must use multi-variable calculus, however, let the three points be designated by their angular position on a circle of radius $2$ (this is for convenience; the radius of the circle does not matter).  Without loss of generality, let the first point $A$ be given angle $0$; the other points $B$ and $C$ are at angles $\alpha_1$ and $\alpha_1+\alpha_2$, with both $\alpha_1, \alpha_2 > 0$ and $\alpha_1+\alpha_2 < 2\pi$.
We can then subdivide the triangle into three sub-triangles: With the center of the circle at $O$, we have $\triangle OAB, \triangle OBC, \triangle OCA$.  These combine to form $\triangle ABC$.  Do not worry if $\alpha_1$ or $\alpha_2 > \pi$; we will determine such a sub-triangle's area as negative, and you can convince yourself (with a diagram) that everything will work out.
Consider the sub-triangle $\triangle OAB$.  It has base length $OA = 2$, and altitude $2\sin \alpha_1$; therefore, its area is $2\sin \alpha_1$.  Likewise, the other two sub-triangles have area $2\sin \alpha_2$ and $2\sin (2\pi-\alpha_1-\alpha_2) = -2\sin (\alpha_1+\alpha_2)$.  We therefore want to maximize their sum
$$
A = 2[\sin \alpha_1+\sin \alpha_2-\sin (\alpha_1+\alpha_2)]
$$
with respect to $\alpha_1$ and $\alpha_2$ (subject to the constraints above).  Can you take it from here?
A: I'm sure this can be done with multivariable calculus, but I would just do the following.


*

*Fix (for a moment) one side of the triangle. Using that side as a base show that the height is maximized, when the other two sides have equal length. This is a necessary condition for maximal area. Remember that the height is the projection of the third vertex of the triangle on the line perpendicular to the base, passing through its midpoint.

*Do the same for all the sides of the triangle, and rejoice.

