# Proof that for increasing AC MC(Q)>AC(Q) at any Q>0? [closed]

I know the "common sense" proof, but how can we prove it algebraically?

## closed as off-topic by luchonacho, emeryville, Herr K., Giskard, Adam BaileyOct 3 '17 at 9:12

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Let $C(y)$ denote the cost function and $y$ the quantity.

The average cost function is $AC=C/y$. This is increasing if its derivative is positive. Let $C'$ denote the derivative of $C$. The marginal cost function is $MC=C'$. We have:

$\frac{\partial AC}{\partial y}= \frac{C'y - C}{y^2}>0$ by using the quotient rule. As $y^2>0$, the condition becomes:

$C'y - C>0$.

Rearranging yields:

$C' > C/y$

Hence: $MC>AC$.

We can calculate total cost as TC(Q)= Q*AC(Q) now just take derivative on both sides w.r.t Q so we get MC(Q)= AC(Q)+ Q * derivative of average cost w.r.t Q. Since Average cost is increasing function derivative must be non-negative so Marginal cost always greater than average cost.