# How to Graph Short-Run Average Cost?

I've learnt to roughly draw graphs of various functions like isoquants of Cobb Douglas function, i.e., $k=√q/L$. Here first derivative is negative so it's downward sloping and second derivative is positive so convex to origin. Now if Short-Run cost function is $C = (w/k)q^2 + (rk)$ then average cost is $AVC= (w/k)q +(rk)/q$. First derivative is $(w/k)-(rk)/q^2$ but how do I know if it's positive or negative?

• Hint: solve for $q$ in $(w/k)-(rk)/q^2>0$ to get the values of $q$ for which $AVC$ is upward sloping, and solve for $q$ in $(w/k)-(rk)/q^2<0$ for values of $q$ that correspond to the downward sloping part of the $AVC$ curve. – Herr K. Sep 23 '17 at 19:26
• Okay so slope changes at $q = √(r/w)*K$ . Now the original graph shows it's oblique asymptotic. How do I find that? – Shoaib Ashraf Sep 24 '17 at 3:18


For the sake of simplicity, call

$$f(q) = \frac{w}{k}q + \frac{rk}{q} = rk\left(\underbrace{\frac{1}{q}}_{\fone} + \underbrace{\frac{w}{rk^2}q}_{\ftwo} \right) = rk (\fone + \ftwo)$$

where I have factored $rk$ out of the expression. Now you want to understand each term separately:

$\fone = 1/q$

This term is drops as $q$ increases, and diverges when $q$ is small.

$\ftwo = \alpha q$, with $\alpha = w/rk^2$

This is a linear term with slope $\alpha$: it is small for small $q$ and large for large $q$.

Combined

In this particular case the function one of the terms grows while the other shrinks. So in extreme cases only one matter. The question is where is the point in which one becomes more relevant than the other.

If you notice above I always use the expressions small and large, but these are relative words. You can actually find a value $q^*$ at which these two terms are equal, and this defines in which each term dominates.

So, if $q < q^*$ this is what I mean by small $q$ and therefore $\fone$ dominates. If, on the other hand $q > q^*$ $f$ will be dominated by $\ftwo$. To find $q^*$ we make

\begin{eqnarray} \fone &=& \ftwo \\ \frac{1}{q} &=& \alpha q \\ q^* &=& \alpha^{-1/2} \end{eqnarray}

With this in mind, below there's a graph for $\alpha = 1$