Maximum distance of a point from a triangle

Question

Consider the triangle in the plane whose vertices are in $A=(0,1)$, $B=(0,0)$ and $C=(2,0)$. Distance of the point $P=(-1,0)$ from the triangle cannot happen if the target point $T$ is on $AB$ or $BC$ (excluding the limit point $B$) because geometrically points on $AC$ has more distance to $P$. Consider a point $Q$ on $BC$ and let $x$ be the length of $CQ$. Then square of $PT$ is $PT^2 = TQ^2 + PQ^2 = (\frac12 x)^2 + (3-x)^2$ and differentiating $PT^2$ and equating to zero gives $x=2.4$ which is geometrically wrong! The book say answer is $T=C$ that is in my solution I must reach at $x=0$ but I couldn't. Please help!

Answer 1

Let $T \in AC$ then $T = tA+ (1-t) C = (2-2t, t)$ where $t\in [0,1]$. Hence the distance (squared) from $P$ is $$ f(t) = \|T-P\|^2 = (3-2t)^2 + t^2 $$ Then, $f'(t) = 2(3-2t)(-2) + 2t = -12 + 10t$ which implies that $f$ is strictly decreasing for $t\in [0,1]$. It follows that its maximum is at $t=0$. Therefore, $T=C$ and $f(0) = 9$.

For the other two segments you can do the same and compare the 3 maximal distances to conclude that the point $C$ is the desired one.