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C'(t) = \frac{(3)(t^2 + 4) - 3t(2t)}{(t^2 + 4)^2} = \frac{3t^2 + 12 - 6t^2}{(t^2 + 4)^2} = \frac{-3t^2 + 12}{(t^2 + 4)^2}.
Set $ C'(t) = 0 $:
-3t^2 + 12 = 0 \Rightarrow t^2 = 4 \Rightarrow t = 2 \quad (\text{since } t \geq 0).

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