use calculus to maximize the area of a triangle
What is the maximum area of a triangle that can be inscribed in a circle with radius of $5$ units. Solve using calculus. Is there a way to use $A = 0.5 \times b \times h$ and take the derivative?
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$\begingroup$Let the radius be $r$ for now. Suppose that side $BC$ of our triangle $ABC$ has length $x$.
Then if we consider orienting the circle such that $BC$ is horizontal and on the bottom half, then point $A$ should be at the top of the circle, in order to maximize the area by maximizing the height.
Thus we only really need to consider isosceles triangles. Let $O$ be the center of the circle and consider $OA$, $OB$, and $OC$ which divide the triangle into three smaller ones. By the area formula $ab\sin(C)/2$, these triangles have area $r^2\sin(\angle AOB)/2$, $r^2\sin(\angle BOC)/2$, and $r^2\sin(\angle COA)/2$. If we let \angle $BOC$ be $2\theta$, then $\angle AOC = \angle AOB = \pi - \theta$.
Hence the area of $ABC$ is the sum of the three areas which is $$r^2(\sin(2\theta) + 2\sin(\pi-\theta)) = r^2(\sin(2\theta) + 2\sin(\theta)).$$
This is a single variable function that we can nicely maximize by taking a derivative. It is fairly straightforward to show that the maximum occurs when $\theta = 60$, or the triangle is equilateral, so the area becomes $3\sqrt{3}r^2/4$.
In your case the maximum area is $75\sqrt{3}/4$.
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