2
$\begingroup$

In my textbook, the shutdown point is defined as the intersection between the average variable costs curve and the marginal costs curve. Suppose Company A has total costs function $TC(q_A)=\frac{1}{2}q_A^2 + 6q_A + 10$. Then its marginal cost is $q_A + 6$ and its average variable cost is $\frac{1}{2}q_A + 6$. Setting those equal I only get zero, does that mean there is no shutdown point? This is part of a bigger exercise where the author says to pay attention to the shutdown points so I imagine it's not zero. What gives?

$\endgroup$
3
  • $\begingroup$ No problems, thanks for answering regardless. $\endgroup$
    – Saegusa
    Aug 28 '20 at 14:01
  • 1
    $\begingroup$ Doesn't your textbook relate the shutdown point to revenue as well as costs? $\endgroup$ Aug 28 '20 at 15:00
  • $\begingroup$ It does, but if it's mathematically true that the shutdown point is equal to that intersection, does that mean the shutdown point is only at zero in this context? $\endgroup$
    – Saegusa
    Aug 28 '20 at 15:01
3
$\begingroup$

I think that potentially your teacher made a mistake (unless this is some tricky 'gotcha' problem) or if the problem involved actually creating the total cost function from some piece of text there might have been mistake in creating the total cost function. If you dont think that applies then read on.

1. Shut down point is at q=0

The first possibility is that indeed shut down point is simply zero. The shut down point is the point at which average variable cost ($AVC$) reaches its minimum - the minimum point can be either found by calculus (by minimizing the $AVC$ function) or indeed by equating $AVC$ to marginal costs MC $AVC=MC$.

The intuition for this is that normally the average cost curve will always decline when $AVC>MC$ and starts increasing only when $AVC<MC$ because marginal cost is the cost for the last product and intuitively if the last added variable to the sample is higher than average of sample then addition of that new variable increases the whole average and vice versa. As a consequence the minimum of average cost curve will be reached when $MC=AVC$ See the example below from Mankiw principles of economics:

enter image description here

However, in your case marginal cost is always above average variable cost ($MC>AVC$) except when $q=0$ when they are actually equal. You have a special total cost function which has average cost function and marginal cost function that looks nothing like the one in textbook. In fact I decided to model your cost function in R in order to drive the point. This is how your marginal cost (in blue) and average cost (red) look like:

enter image description here

As you can see in your case the marginal cost (blue line) is always above average variable costs (red line) except for when $q=0$.

2. Could your teacher possibly mean firm exit (long-run shutdown)?

It is not uncommon for teachers to call exit from the market long-run shut down (see one example here from Texas tech university lecture notes). In the case your teacher meant exit as a long-run shut down then long run shut down point would be given at an intersection of average total cost and marginal costs (see below example from Mankiw's principles of economics).

enter image description here

In this case $ATC=MC$ would equate at some non-zero $q$

3. The above only gives you $q$ coordinate of shut down or exit point

Note that in both examples provided above for firm to actually shut down or exit the price needs to be below the minimum AVC (shut down) or ATC (exit) respectively. The shut down quantity $q_{\text{shut down}}$ only tells you the $x$ coordinate at which AVC reaches its minimum. To see how firm will actually behave you need to always check whether the price (which will be on $y$ axis) that correspond to the shut down coordinate is actually lower than either minimum point on AVC (in case of shut down) or ATC (in case of exit) as shown in the pictures from the Mankiw's textbook.

$\endgroup$
1
  • 1
    $\begingroup$ Thank you for your very detailed answer! I think the confusion arose from not noticing point 3 above. The way he explained the question was somewhat misleading. I think what he meant was not so much $q_{shut down}$ but rather $p(q=0)$, as you're correctly pointed out. In this case, if $P<6$ then the firm doesn't produce. This is most likely what he meant by "shutdown point" in this particular exercise. Thanks for the very clear answer however, it's helped me sort out the question. $\endgroup$
    – Saegusa
    Aug 28 '20 at 15:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.