Max +min of y

For variety, another solution:

$\begin{aligned} y & = \sqrt{(x-2)(7-x)} = \sqrt{-x^2+9x-14} = \sqrt{\frac {25}4-\left(x-\frac 92\right)^2} \end{aligned}$

For $y$ to be real, $\sqrt{\dfrac {25}4-\left(x-\dfrac 92\right)^2} \ge 0$ , when $\left(x-\dfrac 92\right)^2 = \dfrac {25}4$ or $x=2$ and $x=7$ . And $y$ is maximum when $x-\dfrac 92 = 0$ and $\max (y) = \dfrac 52$ . Therefore, $\min(y) + \max(y) = 0 + \dfrac 52 = \boxed{2.5}$ .

By AM-GM, $0 \leq \sqrt{(x-2)(7-x)} \leq \frac{(x-2)+(7-x)}{2} = \frac{5}{2}$ and these bounds are attained, so the answer is $0+\frac{5}{2} = \boxed{2.5}$

It's easy to see that the minimum value of $y$ is $0$ since the function is real-valued and $x=2$ and $x=7$ are in the domain. To maximize $\sqrt{-x^2 + 9x - 14}$ we can simply maximize $y^2 = -x^2 + 9x - 14$ which is going to be the maximum at the vertex in:

$y^2 = \frac{-(b^2 - 4ac)}{4a} = \frac{25}{4} \Rightarrow y = \frac{5}{2}$