Given that $C' = \large\frac{dC}{dx}$$ = 3 + 0.25x$, Rearranging and integrating, we get:

$C(x) = 3x + 0.25 \large\frac{x^2}{2}$$+k$

Given that $C(0) = 60 \rightarrow 60 = 3 \times 0 + 0.25 \large\frac{0^2}{2}$$ + k \rightarrow k = 60$

Therefore, the total cost function is given by $C(x) = 3x + 0.125x^2 + 60$