Cost Minimization with Three Variables [Microeconomics I Studied #35]

OK so once we throw a fixed cost into the mix, we can analyze a slightly more elaborate setup.

(Honestly? I think this whole “more elaborate” version is kinda pointless. The principle is literally the same. (sigh))

Up till now, we’ve been assuming the firm’s inputs are just L and K. This time, let’s say the firm also uses some materials M on top of L and K.

Call the per-unit material price m. Then total cost looks like

$$TC \quad = \quad wL \quad + \quad rK \quad + \quad mM$$

But — this is short-run analysis, so one of these has to be locked in as a fixed cost.

Hmm… let’s fix K at $\overline{K}$.

OK so the reason I said this is pointless — here we go.

Think about the isoquant.

When we had 2 variables, L and K, the isoquant came out as a (curvy) line. So with 3 variables, wouldn’t the isoquant come out as a (curvy) surface?

Like, when

$$Q \quad = \quad AL^{\alpha}K^{\beta}$$

it looked like this [image],

so when

$$Q \quad = \quad AL^{\alpha}K^{\beta}M^{\gamma}$$

wouldn’t it look something like this [image]?????

And the isocost line works the same way. With two variables it was a straight line, so in three dimensions it’d be a flat plane like the shape below.

[image]

So the cost-minimizing point is

[image]

the point where the isocost-plane and the isoquant are tangent — just some single point.

That point’s $(L, K, M)$ is exactly the $L, K, M$ that minimize cost.

But! Once we plug in the assumption that K is fixed at $\overline{K}$,

then in every ordered tuple, K is $\overline{K}$,

[image]

which means we’re only allowed to analyze on this one plane.

So what do the intersection lines on that plane look like?

[image]

The thick black line is the points of the isoquant where $K = \overline{K}$, and the thick red line is the points of the iso-cost line where $K = \overline{K}$.

So out of L, K, M — we just treat K as $\overline{K}$ and only decide L and M.

In other words, it collapses to a problem on coordinate axes like this

[image]

and we’re back to a 2D problem. From here on, same playbook we’ve been using all along.

We can also see this algebraically — it really does reduce to 2D.

Given

$$Q \quad = \quad AL^{\alpha}K^{\beta}M^{\gamma}$$

since K is fixed at $\overline{K}$,

$$Q \quad = \quad A\overline{K}^{\beta}L^{\alpha}M^{\gamma}$$

we can rewrite it like that, and since the isoquant is just the set of points where Q is constant,

$$\frac{Q}{A\overline{K}^{\beta}} \quad = \quad L^{\alpha}M^{\gamma}$$

if we move things around like this, the left side is a constant and the right side is the variable part.

How is this any different from

$$Q \quad = \quad AL^{\alpha}K^{\beta}$$

…..

$$\frac{Q}{A\overline{K}^{\beta}} \quad = \quad L^{\alpha}M^{\gamma}$$

we can just rename the left-hand constant and write

$$Q' \quad = \quad L^{\alpha}M^{\gamma}$$

and that’s that.

Same story for the iso-cost line.

$$TC \quad = \quad wL \quad + \quad rK \quad + \quad mM$$

Set K to the constant $\overline{K}$, and since the iso-cost line is the set of points with “the same cost,” TC in this equation is also a constant.

So,

$$TC \quad - \quad r\overline{K} \quad = \quad wL \quad + \quad mM \quad \text{and writing this as} \\ TC' \quad = \quad wL \quad + \quad mM$$

how is this any different from the 2-variable case? It’s all the same.

$$Q' \quad = \quad L^{\alpha}M^{\gamma}$$$$TC' \quad = \quad wL \quad + \quad mM$$

It becomes the problem of finding the optimum (the tangent intersection) of two curves.

That’s the principle. OK, let’s actually solve one example.

[image]

Originally written in Korean on my Naver blog (2016-07). Translated to English for gdpark.blog.

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