
Re: Trigonometric area optimization
Posted:
Dec 16, 2012 11:55 AM


> ... I seem to come up with m = 3 (not > exactly 3 maybe, but darned close).  1/3 is the > slope of the other line.
Hi Thor, bobmck and Peter,
In my posted reply of 11 Dec I treated the problem from First Principles, because that is the best way to understand a problem.
Since then bobmck has suggested a neat quick solution using a powerful, easily proved, general theorem on the minimum area of a triangle formed by a vertex angle (not necessarily a right angle) at a given point and a fixed line not thru the point. The triangle of minimum area is isosceles wrto the vertex.
Accepting the above theorem, we can proceed as follows:
It was shown that OA = sqrt(1+m^2).10/(m+2) and OB = sqrt(1+m^2).10/(2.m1)
The required minimum triangle has OA = OB .: m+2 = 2.m1 so m=3 QED.
Regards, Peter Scales.

