Originally Posted by
danny arrigo Sorry, when I said it (**) was correct it isn't (should be 4000 instead of 400). Let me provide some details
From $\displaystyle C = 10 l + 20\sqrt{(100-l)^2 + 30^2}$
then
$\displaystyle C' = 10 - \frac{20(100-l)}{\sqrt{(100-l)^2+900}}$
Setting this equal to zero gives
$\displaystyle
2(100-l) = \sqrt{(100-l)^2+900}\;\; \Rightarrow\;\;4(100-l)^2 = (100-l)^2 + 900
$ so $\displaystyle 100-l = \pm \frac{30}{\sqrt{3}}$
Thus $\displaystyle l = 100 \pm 10\sqrt{3}$ where we choose the negative case as the postive gives to much.