Dynamic Geometry: P42

The area of a $\triangle ABC$ is given by $A_\triangle = \dfrac {ab \sin C}2$ . Since side lengths $a=8$ and $b=15$ are fixed, $A_\triangle$ is maximum, when $\sin C = 1$ or $C = 90^\circ$ . Then $c = \sqrt{a^2+b^2} = \sqrt{8^2 + 15^2}= 17$ , which is within $7$ and $23$ . The circumradius $R$ is given by $2R = \dfrac c{\sin C} = 17 \implies R = \dfrac {17}2$ . The inradius $r$ is given by $A_\triangle = sr$ , where $s = \dfrac {a+b+c}2 = \dfrac {8+15+17}2 = 20$ is the semiperimeter of the triangle. Therefore, $r = \dfrac {A_\triangle}s = \dfrac {\frac 12 \times 8 \times 15}{20} = 3$ . Therefore $\dfrac Rr = \dfrac {17}6$ and the required answer is $17+6 = \boxed{23}$ .