Two points are given as A(2,0) and B(8,0). What's the value of y (y>0), so that C(0,y) is such that angle ACB has maximum value?

Question

My first guess is that it could be found as first derivative of some function, but I don't have idea what that function could be.

Answer 1

The required angle is given by $$\theta=\arctan\frac y2-\arctan\frac y8$$

When we differentiate and set to zero, we solve $$\frac{2}{4+y^2}=\frac{8}{64+y^2}\Rightarrow y=4$$ Which leads to $$\theta=\arctan2-\arctan\frac 12=\arctan \frac 34$$

Answer 2

we first note that $$AB=6,BC=\sqrt{(64+y^2)},AC=\sqrt{(4+y^2)}$$ then we use the theorem of cosines $$36=4+y^2+64+y^2-2\sqrt{(4+y^2)}\sqrt{(64+y^2)}\cos(\gamma)$$ and from here we get $$\cos(\gamma)=\frac{(16+y^2)}{\sqrt{(4+y^2)}\sqrt{(64+y^2)}}$$ you must differentiate this with respect to $$y$$