(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

Find the dimensions of an isosceles triangle of least area that can be circumscribed about a circle of radius R.

2. Relevant equations

Let a,b, and c denote the lengths of the sides of this triangle and c=b. Then the angles opposite these sides are A, B and C respectively, (B=C). Let B' be the angle between the radius R and the side a. See attached figure.

3. The attempt at a solution

The quantity to be minimumized is the area of the triangle. I am not too sure how to go about this but the area of the triangle [tex] S=\frac{a}{2}(R+R \sin(B'))[/tex]. Now 2B'+2A=pi, therefore [tex] B' =\frac{\pi}{2}-A \mbox{ therefore} \sin (B') = \sin(\frac{\pi}{2}-A) = \sin(\frac{\pi}{2}+A)[/tex]. Therefore [tex]S=\frac{a}{2}R(1+\sin(\frac{\pi}{2}+A))[/tex].

If I differentiate this implicitly I get [tex] \frac{a}{2}+\frac{a}{2}R\frac{d}{dR}R \sin(\frac{\pi}{2}+A) + \frac{a}{2} \sin(\frac{\pi}{2}+A)=0[/tex].

I need to find an expression for the [tex] \sin(\frac{\pi}{2}+A)[/tex]. I cannot see one suitable.

thanks for helping.

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# Homework Help: Applied minimum problem

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