Max-Min problem - solve only with calculus

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What is the maximum possible area for a triangle inscribed in a circle of radius r?

So the triangle must be an isosceles one. How do we know that, and how do we prove that? And then how should we proceed?

Thanks!

the scarlet macaw
 
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Two points on the circle define a chord. A third point defines a triangle. Use the chord as the base and the third point will give the maximum area when it's perpedicular distance from the chord (height of triangle) is a maximum. You should have no trouble determining that the maximum height occurs when the third point is on the perpendicular bisector of the chord - i.e. it is an isosceles triangle.

The rest is trivial using the vertex angle at the third point as your variable.
 
how do you prove the isosceles-ness of the triangle? I am thinking about trying to set dh/d(theta) = 0 somehow, but other variables keep on getting into the way. Sorry, it may seem clear to y'all, but I am very weak in proofs.
 
As I indicated before, you will find the third vertex of the triangle to lie on the perpendicular bisector of the chord. Draw a line from the vertex to the base (chord). That line bisects the chord so you can see two right triangles. Their bases are identical and the line you just drew is common to both. Therefore, the two triangles are congruent so your original triangle is isosceles.
 
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