Short-Run Total, Average, and Marginal Cost Curves (STC, SAC, SMC) [Microeconomics I Studied #39]

Up to last time we covered TC, AC, and MC — and through that whole conversation we casually let L and K roam over any range we liked, no constraints. Yeah. That whole discussion basically assumed the firm could make its L and K decisions flexibly, which makes it the long-run perspective.

So now we flip the camera and look at things from the short-run perspective.

As I mentioned waaay back, the difference between long-run and short-run isn’t really about how much time has passed — it’s about whether you have flexibility in your decision-making. So looking at things from the short run means no flexibility in decisions: one of L and K is locked down.

OK so, like before, let’s lock K. This locked-down K is now a constant, $\overline{K}$ . And this $\overline{K}$ costs $r$ per unit, so the cost coming from $\overline{K}$ is $r\overline{K}$ .

Oho — so the cost $r\overline{K}$ is a constant too, it doesn’t budge (as long as $r$ doesn’t change)!!!! In other words, $r\overline{K}$ is what we can call the Fixed Cost. Taking the initials of Total Fixed Cost, I’ll write TFC = $r\overline{K}$ .

Then there’s the cost burned by L, the variable that can change: $r\overline{K}$ . This cost is a variable (because L changes). Taking the initials of Total Variable Cost, I’ll write TVC = .

Add ’em up and you get the Total Cost (TC)!!! But to keep it apart from the long-run TC, we’ll call TFC + TVC the STC (Short-run Total Cost).

Now we’re going to draw a curve for this one too, on axes like these:

Empty axes: cost on the vertical axis, Q on the horizontal axis

The logic is the same as before.

The TVC curve, red and concave at first, then blue and convex after a green dotted line. Red note: when output is small, the marginal cost of raising output goes down, so a curve whose dc/dQ keeps shrinking does the trick. Blue note: beyond some Q level marginal cost starts to climb, so the dc/dQ keeps growing

I’m not going to go through that logic in detail here. Honestly, you can just think of it as a cost with only the variable L in it. And if you had to pick one main reason it’s drawn this shape, you’d pick “efficiency.” For the details, check out #36, pleeease~~~~

We also need to draw TFC, so we can mash TVC + TFC together into STC. So let’s draw TFC!!!~ TFC’s easy lol, it’s a constant. Just draw one constant function.

TFC as a flat line at height rK-bar. Note: whatever output you decide on, the fixed cost stays put like this ^^

Now let’s lump the two functions together.

TVC, red and then blue, and the flat TFC line at height rK-bar on the same axes

Now, shall we draw the graph of the two functions’ sum — i.e. STC??!

For the sum, it looks like we can just slide the TVC function up along the Cost axis by $r\overline{K}$ !!!

STC: the TVC curve lifted by rK-bar, so it starts on the TFC line instead of at the origin

Back when we drew TC, did I ever make a big deal about it passing through (0,0)???? I did. The reason I hammered on it:

STC, no matter what, does NOT pass through (0,0)!!!@@

What that means: “even if you produce 0 units, in the short run there are still costs being burned”!!!!

OK, right about now you might be having a small mental breakdown.

Wait?!?! The shape of TC was!!!

The long-run total cost curve TC, starting at the origin

like this, you said — and that’s the long-run perspective, you said — and now we’re switching to the short-run perspective, and the short-run STC is

The short-run total cost curve: the TVC curve lifted by rK-bar, starting on the TFC line

shaped like that?!?!

Huh?? Then if you just take the long-run curve and shift it up the y-axis by $r\overline{K}$ , doesn’t that give you STC??

It’s an easy thing to misunderstand. But it’s clearly a misunderstanding.

Why? Because when we explained the two graphs, we never once pinned down the functions algebraically. We just sketched the general shape. And more fundamentally — the long-run analysis traced the set of costs that hit the real minimum, while the short-run analysis traced the set of costs that hit the possible minimum (not the real minimum). (See #34 for this.)

So what are we doing now? Here too, we won’t derive an exact algebraic form for the functions. We’ll just upgrade and refresh the general shape. We’re going to lump long-run and short-run together, and figure out exactly how they lump together.

What we know: TC

The long-run total cost curve TC, starting at the origin

looks like this, and STC

The short-run total cost curve STC, starting above the origin on the cost axis

looks like this. Those two facts are literally all we’ve got. So if we lump them into one,

TC in black with many possible red STC curves above it

it’s gotta be one of the many, many possible cases of that red line.

Once we lump them… spoiler, the conclusion first: it ends up drawn like this.

STC in red touching TC in black at a single point and lying above it everywhere else

(All we know is the general shape, so it’s not at all… OK fine, it’s a little different from the pictures above. But notice the general shape hasn’t changed.)

This is how we get to see the point: the two must touch at one point.

The logic for nailing down that point — it’s the expansion path. The short-run and long-run expansion paths were drawn like this:

K–L plane: the blue short-run expansion path, a horizontal line at K-bar, and the red long-run expansion path through the points where isocost lines touch the isoquants

(See #34.)

First — the expansion path is “the set of (L, K) that achieves the minimum cost.” Now let’s read this graph again!!!!

K–L plane: the isoquant Q_1 touching a dotted isocost line at the point (L_1, K_1)

Say the L and K that hit the minimum cost at output Q_1 are (L_1, K_1) . And say the short-run K is locked at K_1 , and let’s bump output Q up from Q_1 to a higher Q_2 .

The same plane with a higher red isoquant Q_2 touching a red dotted isocost line at (L_2, K_2)

Among all the (L, K) that produce Q_2 , the one that hits the minimum cost is (L_2, K_2) . But in the short run, K can’t change to K_2 , because it has to stay locked at K_1 . Meaning — we’ve gotta use K_1 to crank out Q_2 .

With K stuck at K_1, the blue point on Q_2 sits at L’_2, and the blue dotted isocost line through it lies above the red one

That’s the spot where we have to use (L'_2, K_2) , but — the isocost line through sits higher than the red isocost line, as you can see there. So if you use K_1 to produce Q_2 , you burn more cost than the red dotted line.

Same logic — let’s go the other way and cut output from Q_1 down to Q_0 .

A lower red isoquant Q_0 touching a red dotted isocost line at (L_0, K_0), below the Q_1 isoquant

Among the (L, K) that produce Q_0 , the cost-minimizing one is (L_0, K_0) .

With K stuck at K_1, the blue point on Q_0 sits at L’_0, and the blue dotted isocost line through it lies above the red one

But same deal here — in the short run, K can’t change to K_0 , because it has to stay locked at K_1 . Meaning — we’ve gotta use K_1 to crank out Q_0 .

So we end up having to use (L'_0, K_1) , but….. the isocost line through ;;; sits higher than the red isocost line too, as you can see there. So if you use K_1 to produce Q_0 , you also burn more cost than the red dotted line.

OK, back to the cost curves. Let’s stack up what we’ve figured out.

The long-run total cost curve (TC) —

The long-run total cost curve TC, starting at the origin

we know it looks like this. And the general shape of the short-run total cost curve (STC) —

The short-run total cost curve STC, starting above the origin on the cost axis

we know it looks like this. And we found that at one Q point, Q_1 , TC and STC are equal, and for every Q other than that one, $TC \lt STC$ holds.

Now let’s use these facts to fit the pieces together one by one.

STC in red touching TC at Q_1 and pushed above it everywhere else, starting at height rK-bar, with six numbered notes: TC has this general shape; STC is drawn in red and equals TC at Q_1; at every other Q, STC is greater than TC; STC has this shape (a small sketch); drawing STC from all that; its height is rK-bar

Just follow the colors ^^ Make sense…? hehehe

So now, for

STC in red above TC in black

let’s dig into the function for its slope too. For the long run — i.e. the slope analysis with respect to TC — we already did that earlier.

(See #36.)

Here we’ll zoom in more on these, for STC:

STC over Q equals SAC, and the derivative of STC with respect to Q equals SMC

(Short-run Average and Marginal Cost).

OK then: how do SAC and SMC get drawn? Let’s get their general shape first.

STC with dashed rays from the origin to points on it; the solid green ray just touches STC above Q_A

For SAC, it’s the slope from the origin, and there’s clearly a point where SAC bottoms out ‘right there’!!! Let’s call it $Q_A$.

(Oh?! So is this the Q where STC touches TC? Only if the long-run AC bottoms out right there too; in general, no. (whisper whisper))

STC with tangent lines: the slope falls until the solid blue tangent above Q_B and rises after it; the green ray from the origin touches STC above Q_A

Now SMC. Until it reaches the slope of that solid line way over there, the slope of the tangent to STC keeps dropping. Then it gradually starts climbing!!! Let’s call the Q with the minimum SMC $Q_B$!!!!

(Oh!!! At $Q_A$, SMC and SAC must be equal….. (whisper whisper))

With all that info, if we toss SAC and SMC onto the same set of axes,

SMC in blue and SAC in green against Q: SMC bottoms out first and then rises through the lowest point of SAC

Whoa!!!

Deriving AC and MC from TC and deriving SAC and SMC from STC give exaaaactly the same general shape?!?!

OK so now, for real, let’s draw AC, MC, SAC, and SMC all lumped together. This has already gone way too long, so I’ll jump straight into the next post and pick it up there — no yakking.

Heads-up. The figures here are the original notes, slips and all. A few things to watch for:

The cost of the variable input $L$ is $wL$; the small picture right after "the variable that can change" shows $r\overline{K}$, the fixed cost, by mistake.
In the short-run passage, the point we are stuck with is $(L'_2, K_1)$, since capital stays at $K_1$; the two small pictures write $(L'_2, K_2)$.

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