What is the maximum perimeter of a triangle inscibed in a circle of radius $1$?
I can't seem to find a proper equation to calculate the derivative.
[Math] Max perimeter of triangle inscribed in a circle
derivativesgeometryoptimization
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I thought an image could help:
The perimeter is $$\overline{BC} + 2\overline{AB} = 2\sin \alpha + 2\cdot 2\sin \beta = 2\sin \alpha + 2\cdot 2\sin \left(\frac{\pi}{2} - \frac{\alpha}{2}\right) = 2\left(\sin \alpha+2\cos\left(\frac{\alpha}{2} \right) \right)$$ For further simplifications let $\alpha = 2t$, then you have to maximize $$f(t) = 2\left(\sin 2t+2\cos t) \right) = 2(2\sin t \cos t + 2\cos t ) = 4(\sin t \cos t + \cos t) $$ Finding $f'(t)$ is easy: $f'(t) = 4(\cos t \cdot \cos t + \sin t \cdot -\sin t -\sin t) = 4(\cos^2 t - \sin^2 t -\sin t )=4(-2\sin^2 t -\sin t +1)$ and it equals $0$ when $\sin t = \frac12 \Rightarrow t = \frac{\pi}{6} \Rightarrow \alpha = 2t = \frac{\pi}{3} $ so the triangle is equilater!
Computing the area and perimeter, we get $A=x\sqrt{1-x^2}$ and $P=2+2x$.
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Therefore, $$ \begin{align} r^2 &=\frac{4A^2}{P^2}\\ &=x^2\frac{1-x}{1+x}\tag{1} \end{align} $$ This reaches maximum when the derivative of $(1)$ is $0$; that is, when $$ x^3+x^2-x=0\require{cancel}\tag{2} $$ The only root of $(2)$ within the allowable range of $x\in(0,1)$ is $x=\frac{-1+\sqrt5}2=\frac1\phi$. Thus, the maximum area of the inscribed circle is $$ \begin{align} \pi r^2 &=\frac\pi{\phi^5}\\ &\doteq0.283277232857953\tag{3} \end{align} $$ Divide the area by $\pi$ and take the square root to get the radius of the circle.
Best Answer
HINT:
Use law of sines and this