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.Com
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) (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 \ =\ \frac{\tan\alpha\ -\ \tan\beta}{1\ +\ \tan\alpha\,\tan\beta})
, you should be able to derive the fact that:
\ =\ \frac{t\ +\ 3}{t^2\ +\ 1})
Use the quotient rule to find the derivative,
\right)}{dt}\ =\ \frac{-t^2\ -\ 6t\ +\ 1}{\left(t^2\ +\ 1\right)^2})
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
Egw to Beta kai to Sigma
My calculator said it, I believe it, that settles it
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