Given the assumption that $C(q)$ is continuously differentiable we have for all $q$: $$qAC(q) = C( q ) = FC + \int_0^{q} MC(x) \text{d}x.$$ Taking the difference for any pair $q,\hat{q}$: $$qAC(q) - \hat{q}AC(\hat{q}) = \int_{\hat{q}}^{q} MC(x) \text{d}x. \tag{1}$$ The left hand side may be reformulated: $$q\left(AC(q) - AC(\hat{q}) \right) + \left(q - ... 2 You claim that AC is minimised at MC=AC thus at this quantity q_c we have AC'(q_c) = 0. We will show that given the assumptions MC'(q_c) \geq 0, that is MC cannot be decreasing in q at this location. For all q it is true that$$ AC'(q) = \frac{MC(q)-AC(q)}{q}. $$For all q > q_c we have$$ 0 = AC'(q_c) = \frac{MC(q_c) - AC(q_c)}{q_c} = \...