LC oscillator conservation of energy

The oscillator circuit can be modeled via KVL according to:

$0 = Lq''(t) + \frac{1}{C}q'(t), q(0)=10, q'(0)=0$ (i)

which has the general solution:

$q(t) = A\cos(t/\sqrt{LC}) + B\sin(t/\sqrt{LC})$ (ii)

and after applying the boundary conditions in (i) to (ii), we finally obtain:

$q(t) = 10\cos(t/\sqrt{LC})$ (iii),

$i(t) = q'(t) = (-10/\sqrt{LC})\sin(t/\sqrt{LC})$ (iv).

From (iv), $\frac{i_{MAX}}{2} = \frac{5}{\sqrt{LC}} \Rightarrow \frac{5}{\sqrt{LC}} = (-10/\sqrt{LC})\sin(t/\sqrt{LC}) \Rightarrow t/\sqrt{LC} = \arcsin(-1/2) = -\pi/6$ . Thus, the capacitor's charge at this instance of time computes to $q = 10\cos(-\pi/6) = 10(\sqrt{3}/2) = \boxed{5\sqrt{3}}$ coloumbs.