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1

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} = \...


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