SOLUTION: A(1,2), B(-1,1), C(1,0), D(-1,0) and P is (t,0). (cordinates of p are located between 0 and 1 on axis X), we are to find the point P at which angle APB is maximized tan(APC) is

Algebra ->  Triangles -> SOLUTION: A(1,2), B(-1,1), C(1,0), D(-1,0) and P is (t,0). (cordinates of p are located between 0 and 1 on axis X), we are to find the point P at which angle APB is maximized tan(APC) is       Log On


   



Question 746890: A(1,2), B(-1,1), C(1,0), D(-1,0) and P is (t,0). (cordinates of p are located between 0 and 1 on axis X), we are to find the point P at which angle APB is maximized
tan(APC) is = 2/1-t, tan(BPD) = 1/1+t and hence tan(APC) = t+x/t^2+y
therefore, the cordinates of point P are (t,0)

Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!


Let refer to angle APC. Let refer to angle APB. Our task is to maximize

is indeed , but you have a sign error or . Should be .

Using those values, and the fact that , you should be able to derive the fact that:



Use the quotient rule to find the derivative,



Set the numerator of the derivative equal to zero and solve the quadratic for the positive value for that yields a maximum tangent, and therefore a maximum angle.

John

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