Hi Adam, > > The lines y=10-2x, 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=10-2.x be AB
Then Area =1/2 .OA .OB
A is 10/(m+2),m.10/(m+2) B is 10/(2-1/m),10/(1-2.m)
Area=1/2 .OA .OB =50.(1+m^2)/(2.m^2+3.m-2) 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.