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$\displaystyle 15 = 2L + (\pi+2)r$
$\displaystyle L = \frac{15 - (\pi+2)r}{2}$
$\displaystyle A = 2rL + \frac{\pi r^2}{2}$
$\displaystyle A = 15r - (\pi+2)r^2 + \frac{\pi r^2}{2}$
$\displaystyle A = 15r - \left(\frac{\pi}{2} + 2\right)r^2$
$\displaystyle \frac{dA}{dr} = 15 - (\pi + 4)r = 0$
$\displaystyle r = \frac{15}{\pi + 4}$
I get $\displaystyle \frac{r}{L} = 1$