
Re: Trigonometric area optimization
Posted:
Dec 14, 2012 10:58 PM


Let me just add that you can prove the isoceles right triangle just spoken of has the minimum area. I turned it into an angular problem and minimized the function expressed in terms of theta, the angle that one of the legs of the right triange you spoke of makes with the hypotenus. The area can be found in terms of a product of sines and cosines of that angle, and the magnitude of the legs. putting the derivative equal to zero, I finally got sine squared theta equals cosine squared theta, which happens at forty five degrees, the condition that creates an isoceles right triangle.

