
Re: Trigonometric area optimization
Posted:
Dec 11, 2012 12:25 PM


Hi Adam, > > The lines y=102x, y=mx and y=(1/m)x, where m > 1/2, > form a right triangle. Find an m, so that the area > measurement of the triangle is as small as possible. > > Let line y=m.x be OA, line y=(1/m).x be OB and line y=102.x be AB
Then Area =1/2 .OA .OB
A is 10/(m+2),m.10/(m+2) B is 10/(21/m),10/(12.m)
OA=sqrt(1+m^2).10/(m+2) OB=sqrt(1+m^2).10/(2.m1)
Area=1/2 .OA .OB =50.(1+m^2)/(2.m^2+3.m2) Find dArea/dm and set it to = 0 and select the option with m>1/2
Sketch the result and check Area for m=0 and a few other values of m, including both solutions of the quadratic dArea/dm=0.
Regards, Peter Scales.

