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\begin{document}

 
 
 
%\vspace*{12pt}

 
\noindent
October 11, 2018\\
{\Large {\bf Notes on the          Shutdown Decision }} 
 
 
   

\noindent
{\sc 1. Teaching the Decision to Shut Down if $P<Min(AVC)$}

The decision to shut down if $P<Min(AVC)$ is hard to teach and hard to learn. Below I've written up two numerical examples that might help.  The first is a cubic equation. Someone could improve it by making it work out to more even numbers, perhaps. The second is a piecewise equation with a V-shaped AVC and a discontinuous MC (a kinked TC curve). The numbers work out to be even, but the piecewiseness is kludgy and the discontinuity will perturb students--- though it would actually be a good thing for helping them to understand the intuition. 


 I think what might be even better would be a discrete numerical example that would allow more of a storytelling approach. 

\newpage

 
 \noindent
{\sc 2. U-Shaped Marginal Costs}

    \begin{center}
      {\sc    Figure 1:\\
         The Total Cost Curve  $TC =144 + 32Q - 8Q^2 +Q^3$ } \label{shutdown1}
          
  \includegraphics[width=3in]{shutdown1.jpg}
  
 
      {\sc    Figure 2:\\
         The Average  Total Cost, Average Variable Cost, and Marginal Cost Curves } \label{shutdown2}
          
  \includegraphics[width=3in]{shutdown2.jpg}
   
      \end{center}

Suppose the total cost is 
   \begin{equation} \label{e0} 
  {\rm Total\; Cost}  =144 + 32Q - 8Q^2 +Q^3
\end{equation}
 

 The average total cost is  $\frac{  {\rm Total\;  Cost} }{Q}$, so it is 
  \begin{equation} \label{e0} 
  \begin{array}{lll l}
  {\rm Average\; Total\;Cost} & =&\frac{144}{Q} + 32- 8Q + Q^2  
  \end{array}
\end{equation}

   The minimum of the average total cost is found by differentiating it with respect to $Q$ and setting the derivative equal to zero--- finding the ``bottom of the valley''. Doing that we get, since  \\
    $ATC= \frac{144}{Q} + 32  -8Q + Q^2$ means that \\
    $ATC = 144Q^{-1} +32  -8Q + Q^2$,   
    \begin{equation} \label{e0} 
  \begin{array}{lll  }
  \frac{d {\rm ATCt}} {dQ}&=& (-1)144 Q^{-1-1}  -8 + 2Q  =0  \\
     && \\
    &  = &  \frac{-144}{Q^2} -8 + 2Q=0    \\
   && \\
  &  \rightarrow  & - 8Q^2 + 2Q^3 =144    \\
     &&\\
   &  \rightarrow &   Q^3 - 4Q^2 =72    \\
     &&\\
   &  \rightarrow& Q  =  6  \\
   && \;  {\rm  The\; minimum\; ATC = \frac{144}{6} +32  -8\cdot 6 + 6^2 
    =  24+32 - 48+36 = 44}   \\   
  \end{array}
\end{equation}
   The price will have to rise to $P=44$ for the firm to make a long-run profit, covering its fixed costs as well as its variable costs. 
   

 
The fixed cost is 144, and the variable cost is whatever is left over. Thus
 \begin{equation} \label{e0} 
  \begin{array}{lll l}
  {\rm Average\; Variable\;Cost} & =&  32- 8Q + Q^2 \\
  \end{array}
\end{equation}

   The minimum of the average average cost is also found by differentiating it with respect to $Q$ and setting the derivative equal to zero--- finding the ``bottom of the valley''. Doing that we get 
    \begin{equation} \label{e0} 
  \begin{array}{lll l}
  \frac{d {\rm AVC}}{dQ}   = -8 +2Q =0 && & \\
     &&&\\
   Q  =  4   \;  {\rm  has \; the\; minimum\; AVC = 32- 8\cdot 4 + 4^2= 16}&& & \\   
  \end{array}
\end{equation}
  
  
  
\newpage


 The marginal cost is the derivative of   total cost,\\
  $TC  =144 + 32Q - 8Q^2 +Q^3$,  so it is
 \begin{equation} \label{e0} 
  {\rm Marginal\; Cost} =\frac{d {\rm TC}}{dQ}     =   32  - 16Q  + 3Q^2 
\end{equation}

 The minimum of the marginal cost is also found by differentiating it with respect to $Q$ and setting the derivative equal to zero--- finding the ``bottom of the valley''. Doing that we get 
    \begin{equation} \label{e0} 
  \begin{array}{lll }
  \frac{d {\rm MC}}{dQ}  & =& -16 +6Q =0     \\
     && \\
  &\rightarrow& Q  =  \frac{16}{6} = 2 \frac{2}{3} \\
             && \\
      {\rm with  \;} MC& =& 32- 16(8/3) + 3 \cdot (8/3)^2=  32-128/3+ 3 (64/9) = 32- 128/3 + 64/3 = 96/3 - 64/3 = 32/3\\

       & =& 10 \frac{2}{3}      \\   
  \end{array}
\end{equation}

\newpage

What is the optimal output for the firm? That depends on the market price. If the price is zero, the firm should produce  $Q=0$, for profits of -144. It has to pay the fixed cost of  144, but the marginal cost is 16 at $Q=0$, so it isn't  worth producing anything. Clearly, if $P<10 \frac{2}{3}$, the firm should produce $Q=0$. In Figure 3, for example, the $P=5$ curve shows that profit is always negative, for any $Q$, but it's least negative at $Q=0$, where profit is $-144$, and just gets lower as output is increased. 


\begin{center}
      {\sc    Figure 3:\\
        Profit As a Function of  Output if the Firm Uses the MC(Q) = P Rule Blindly } \label{shutdown2}
          
  \includegraphics[width=5in]{shutdown2a.jpg}
   
      \end{center}
      
 
      

How about if the price rises to $P=10 \frac{2}{3}$, which is $Min(MC)$?  The $MC(Q)=P$ rule is still misleading, because each unit except the last has a marginal cost less than the price.    We'd get $Q=2 \frac{2}{3}$, what we found earlier was the $Q$ that minimized $MC(Q)$, and profit would be $-191$. It's still best to produce $Q=0$. 
 
 How about if the price rises to $P=16$, which is $Min(AVC)$?  The $MC(Q)=P$ rule now starts to work properly.   We'd get $Q=4$, what we found earlier was the $Q$ that minimized $AVC(Q)$, and profit would be $-144$, which is at least just as good as producing $Q=0$.  Notice in Figure 3 that the profit curve is slightly falling starting from $Q=0$, but then is slightly rising up to $Q=4$, after which it starts to fall sharply. The firm should pick $Q=0$ or $Q=4$, but not some output in between. 
 
 
\begin{center}
      {\sc    Figure 3: (repeated)\\
        Profit As a Function of  Output if the Firm Uses the MC(Q) = P Rule Blindly } \label{shutdown2}
          
  \includegraphics[width=5in]{shutdown2a.jpg}
   
      \end{center}
 

   How about if the price rises to $P= 44$?  The $MC(Q)=P$ rule tells us to pick  $Q=6$, what we found earlier was the $Q$ that minimized $ATC(Q)$, and profit would be $0$. Hurrah! We break even. And producing $Q=6$ is much   better than producing $Q=0$. The firm will want to keep in existence even in the long run if we can expect $P=44$ to continue. 

  If $P= 44$, things get even better.   The $MC(Q)=P$ rule tells us to pick  $Q \approx 6.8$, looking at Figure 3,   and profit would be over 100.  

 
 
 Thus, the supply curve should have $Q(P)=0$ if the price is below $10 \frac{2}{3}$  and then it should trace out the MC. We have $P=MC=  32  - 16Q  + 3Q^2$,  so we have $P(Q)=   32  - 16Q  + 3Q^2$, which we need to invert to get $Q(P)$ instead of $P(Q)$.  One (even``you'' if you had   time) can do that with high school's quadratic formula, but I used the computer language Python,\footnote{The Python 2 code is: \\
 from sympy import *\\
from sympy.abc import * \#This makes each letter  and greek letter name   a symbol  \\
answer = solve( 32  - 16*Q  + 3*Q**2-p, Q)\\
print("answer  = ",  answer  , "$\backslash$n") 
} and arrived at  $Q(P) = 2 \frac{2}{3} + \frac{\sqrt{ 3 P - 32}}{3}.  $ The supply equation is therefore
 
  
 \begin{equation} \label{e0} 
  \begin{array}{lll l}
  Q^{supply}(P) & =&  0  & if \;\; P \leq 10 \frac{2}{3}\\                    
    &  & 2 \frac{2}{3} + \frac{\sqrt{ 3 P - 32}}{3} & if \;\;  P \geq 10 \frac{2}{3}. \\ 
  \end{array}
\end{equation}


 \newpage
\noindent
{\sc 3. V-Shaped Marginal Costs}
 
 Suppose the total cost is 
   \begin{equation} \label{e0} 
  \begin{array}{lll l}
  {\rm Total\; Cost} & =&49 + 16Q - Q^2 & if \;\; Q \leq 6\\
  &  &49 + 4Q +Q^2 & if \;\; Q \geq 6\\
  \end{array}
\end{equation}
 

 The average total cost is  $\frac{  {\rm Total\;  Cost} }{Q}$, so it is 
  \begin{equation} \label{e0} 
  \begin{array}{lll l}
  {\rm Average\; Total \;Cost} & =&\frac{49}{Q} + 16  - Q  & if \;\; Q \leq 6\\
  &  &\frac{49}{Q} + 4  +Q  & if \;\; Q \geq 6\\
  \end{array}
\end{equation}
\vspace*{-40pt}
 
       \begin{center}
      {\sc    Figure 3:\\
         The Total Cost Curve, for V-Shaped Average Variable Cost   } \label{shutdown3}
          
  \includegraphics[width=2in]{shutdown3.jpg}
  
 
      {\sc    Figure 4:\\
         The Average  Total Cost, Average Variable Cost, and Marginal Cost Curves } \label{shutdown4}
          
  \includegraphics[width=2in]{shutdown4.jpg}
   
      \end{center}

   The minimum of the average total cost is found by differentiating it with respect to $Q$ and setting the derivative equal to zero--- finding the ``bottom of the valley''. Doing that we get, since the minimum will be at a value greater than $Q=6$ and $ATC= \frac{49}{Q} + 4  +Q = 49Q^{-1} +4+Q$, 
    \begin{equation} \label{e0} 
  \begin{array}{lll l}
  \frac{d {\rm Average\; Total\;Cost}}{dQ}   = (-1) 49 Q^{-1-1} +1  =0 && & \\
     &&&\\
    \frac{d {\rm Average\; Total\;Cost}}{dQ}   =   -\frac{ 49}{Q^2}+1 =0 && & \\
   &&&\\
   49  =  Q^2&&&\\
    &&&\\
   Q  =  \sqrt{49} =7 \;  {\rm  has \; the\; minimum\; ATC = 7+4+7= 18}&& & \\   
  \end{array}
\end{equation}
 
This means the fixed cost is 49, and the variable cost is whatever is left over. Thus
 \begin{equation} \label{e0} 
  \begin{array}{lll l}
  {\rm Average\; Variable\;Cost} & =&  16  - Q  & if \;\; Q \leq 6\\
  &  &  4  +Q  & if \;\; Q \geq 6\\
  \end{array}
\end{equation}

 The average variable cost is V-shaped, and its minimum is 10, at $Q = 6$.  Its minimum is lower than ATC's because it doesn't included the fixed cost. It's at a lower Q  because it doesn't need to spread the fixed cost over more units to get a minimum. 
 
 \bigskip

 The marginal cost is the derivative of   total cost, so it is
 \begin{equation} \label{e0} 
  \begin{array}{lll l}
  {\rm Marginal\; Cost} & =&  16  - 2Q  & if \;\; Q \leq 6\\
  &  &  4  + 2Q  & if \;\; Q > 6\\
  \end{array}
\end{equation}

The marginal cost is V-shaped, and its minimum is 4, at $Q = 6$.  Its minimum is lower than AVC's because it doesn't average in  all of the high  values of marginal cost for $Q<6$. 

Note that the marginal cost jumps at 6, from MC=4 to MC = 16. There is a kink in the total cost curve at $Q=6$, so the slope suddenly increases. 

What is the optimal output for the firm? That depends on the market price. If the price is zero, the firm should produce  $Q=0$, for profits of -49. It has to pay the fixed cost of 49, but the marginal cost is 16 at $Q=0$, so it isn't  worth producing anything. Clearly, if $P<4$, the firm should produce $Q=0$. 

How about if the price rises to $P=4$? Then we could use the rule of picking $Q$ so $MC(Q)=P$. Doing that, we'd have $MC=16-2Q = 4$, so $12=2Q$,  and $Q=6$. But this would be a very bad idea, because $P<Min(AVC)=10$.   The profit at $Q=6$ would be $PQ- TC(Q) = (4)(6) - (49 +16(6) - 6^2) = 24 -49-96+36 = -85$. That's lower than the profit of $-49$ from $Q=0$. It is better if the firm continues to produce $Q=0$. 

That's continues to be true up until the price reaches $P=10$, so $P=Min(AVC)=10$.  At that price, producing $Q=6$ yields a profit of $PQ- TC(Q) = (10)(6) - (49 +16(6) - 6^2) = 24 -49-96+36 = -49$. That's the same profit as from $Q=0$. 

What if the price rises to $P=11$?  The intersection of the MC and market price curves continues to be at $Q=6$. The firm should not increase output beyond $Q=6$, because the marginal cost immediately jumps to $MC=16$. So the elasticity of supply is 0 at $P=11$. In fact, until the price rises to 16, the firm should stick with $Q=6$. 

 If the price rises above 16, though, then at $Q=6$ the marginal cost of 16 would be less than the price, so the firm should start increasing its output beyond 6. If $P=20$, for example, the firm should pick output so $MC(Q) = 4+2Q = P=20$, so $2Q=16$ and $Q=8$. Solving that rule for $P$, we get the supply curve $Q =  \frac{P}{2} -2$ for $P \geq 16$. 
 
 Thus, the supply curve is:
 \begin{equation} \label{e0} 
  \begin{array}{lll l}
  {\rm Q^{supply}} & =&  0  & if \;\; P \leq 10\\
  & &  6  & if \;\;  P \in [10, 16]\\
    &  &  \frac{P}{2} -2 & if \;\;  P \geq 16. \\ 
  \end{array}
\end{equation}
 

 

   


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