Maximum and Minimum value of equations.

$\begin{aligned} P & = \frac {\overline{ab}}{a+b} = \frac {10a+b}{a+b} = 1 + \frac {9a}{a+b} = 1 + \frac 9{1+ \frac ba} \end{aligned}$ . Therefore, $P$ is maximum when $\dfrac ba$ is minimum, that is $\dfrac ba = 0$ , $\implies b = 0$ and $P_{max} = 1 + \dfrac 9{1+0} = 10$ . $P$ is minimum, when $\dfrac ba$ is maximum, that is $\dfrac ba = 9$ , $\implies b = 9$ and $a=1$ , and $P_{min} = 1 + \dfrac 9{1+9} = \dfrac {19}{10}$ .

Therefore, $P_{max} + P_{min} = 10 + \dfrac {19}{10} = \dfrac {119}{10}$ . $\implies x + y = 119+10 = \boxed{129}$ .

We have $P=\frac{10a+b}{a+b}=1+\frac{9a}{a+b}$

We can make this as big as possible by taking $b=0$ , giving $P_{max}=10$ (this is achieved whenever $\overline{ab}$ is a multiple of $10$ ).

Likewise, we make $P$ as small as possible by taking $b=9$ .

We have $P=1+\frac{9a}{a+9}=1+\frac{9a+81-81}{a+9}=10-\frac{81}{a+9}$ .

This last form makes it clear that the minimum is found when $a=1$ (the smallest allowed value), giving $P_{min}=1.9$ when $\overline{ab}=19$ .

The sum of these is $11.9=\frac{119}{10}$ , leading to the answer $\boxed{129}$ .