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.
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Intuitively, the maximum ought to be an equilateral triangle, with perimeter $3\sqrt 3$.
If you want to use calculus, let $\theta$ and $\phi$ be the arcs spanned by two of the sides, and calculate the perimeter as $$ 2\sin \frac{\theta}2 + 2\sin\frac\phi2 + 2\sin\frac{2\pi-\theta-\phi}2 $$ Some manipulation of trigonometric identities will be involved.
hmakholm left over Monica
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Would that require calculating derivative of function of two variables? I don't know how to do that :( – Josh Feb 27 '14 at 14:18
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@Josh: Yes, usually one would do it with partial derivatives. If you don't have those, you might be able to get through by first keeping $\theta$ constant and showing that the maximal perimeter for that $\theta$ occurs when $\phi=\frac{2\pi-\theta}{2}$ (i.e. an isosceles triangle). Then substitute that into the full formula and now differentiate with respect to $\theta$. – hmakholm left over Monica Feb 27 '14 at 14:26
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Hint: the triangle can be defined by two numbers $0 < \theta < \phi < 2 \pi$, where one vertex $A$ say is at (1,0) and vertices $B, C$ are at $(\cos \theta, \sin \theta) \mbox{ and } (\cos \phi, \sin \phi)$.
Use Pythagoras to calculate AB + BC + CA, and differentiate with respect to $\theta, \phi$ to get maximum.
oks
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Would that require calculating derivative of function of two variables? I don't know how to do that :( – Josh Feb 27 '14 at 14:18
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You treat one variable as constant, and differentiate with respect to the other. http://en.wikipedia.org/wiki/Partial_derivative – oks Feb 27 '14 at 14:21