Why is this? Let’s reason about it in a crude way.

It depends on why there is unemployment. It is said that economic policy dictates some buffer stock of unemployed people be maintained to anchor inflation, which is to say that unemployment can be eliminated by inflating. Indeed, at a low enough real interest rate (including negative rate), all work that spans non-zero time becomes “profitable” at wage above zero and there can be no unemployment. So there will be some unstable equilibrium, where there is full employment at just above zero wage and only “unprofitable” work left on the table. For a margin of safety (so as not to overshoot and still leave profitable work at above the lowest wage, which will unanchor the targeted inflation), there will be some unemployment, and therefore always payable yet unprofitable work. If the government offers some delta positive wage, then this excess labor can be used to do the unprofitable work, and we can do away with the technical requirement of excess labor margin and yet get equilibrium without the possibility of overshooting. Effectively, the government inflates the last little bit as needed to pay wages directly … rather than dumping it into the banking system where it will be promptly lent back to the government due to the real interest rate being slightly too high.

Theoretically, this also solves one of the greatest libertarian complaints that the minimum wage causes more unemployment. The theory of minimum wage is that everybody should be able to self-sustain on their wage (which now includes wage from the government as the employer of last resort). Now, by policy, the government can set some minimum wage (but it must be set so low that, it is guaranteed that everybody will definitely spend at least at that wage rate) and pay its own wage at delta below that. Since all of labor now spends at the minimum wage rate, there will be support on the demand side to assure that the full-employment equilibrium is set where the minimum wage cannot be unprofitable. All in all, the government does not compete with the private sector.

Of course there are still practical issues with this. For example, what do you do with lazy workers. So employer of last resort should only be in the sense of employment opportunity of last resort. You should still be fired for substandard work and go without pay at all. Now that’s an incentive to be productive. Another practical issue is churn. Since the government only intervenes at the lower extremum of the wage scale, unemployment higher in wage scale would need to filter down through displacement. I don’t really see a good solution to this. Perhaps there are clever and ethical ways to increase job market liquidity.

Consider an equilateral triangle ABC with edge length 1. At each vertex is an object that is capable of movement at exactly speed 1. Beginning at time 0, each of the three objects moves toward its initial adjacent neighbor object, as in a game of pursuit. Of course, by symmetry, the objects will meet at the incenter of ABC. The question: how far will they have traveled?

The paths followed by the objects at first seem non-trivial, but in fact turn out to be nice. We will come back to this, but for the problem, the exact paths are irrelevant. We just need to find the amount of time it takes the objects to meet. The key part of the solution is to realize that, due to symmetry, the three objects always define an equilateral triangle at any time. Furthermore, their direction of motion is always along the edges of such a triangle, at speed v=1.

What we have is then a shrinking and rotating equilateral triangle that eventually decays to a point. Let us characterize this process.

At time \(t\), let \(s(t)\) be the length traveled by one of the objects. Let \(l(t)\) be the edge length of the triangle. Let \(A(t)\) be the area of the triangle. Then we have several relationships:

• \(\frac{ds}{dt}=1\)
• \(A(t)=\frac{\sqrt{3}}{4} l(t)^2\), and \(\frac{dA}{dt} = \frac{\sqrt{3}}{2} l(t) \frac{dl}{dt}\)
• By geometry, \(A(t+\Delta t) – A(t) = -3 (l(t) – \frac{ds}{dt} \Delta t) \frac{\sqrt{3}}{4} \frac{ds}{dt} \Delta t + o({\Delta t}^2)\), thus \(\frac{dA}{dt} = -3 \frac{\sqrt{3}}{4} l(t)\)

Combining: \(-\frac{3}{4}\sqrt{3} l(t) = \frac{\sqrt{3}}{2} l(t) \frac{dl}{dt}\)
\(\frac{dl}{dt} = -\frac{3}{2}\). Integrating with boundary conditions, we get \(-\frac{3}{2}t_f = -1\), \(t_f = \frac{2}{3}\), and since speed is 1, \(s(t_f) = t_f = \frac{2}{3}\). This is how far the object travels.

Finally, since the process is self-similar at every step along the path, the path of the object must be a logarithmic spiral. Furthermore, since the linear speed on the path is constant, the angular speed must be exponentially increasing (in angle) toward the center of the spiral.

To solve for this spiral explicitly, we have \(\frac{dr}{dt}=\frac{dr}{d\theta} \frac{d\theta}{dt} = \frac{\sqrt{3}}{3} \frac{dl}{dt} = -\frac{\sqrt{3}}{2}\).
\(\frac{dr}{d\theta} = -k \exp(-k \theta)\), so \(\frac{d\theta}{dt} = \frac{\sqrt{3}}{2k} \exp (k\theta)\), and \(\theta(t) = -\frac{1}{k} \ln(-\frac{\sqrt{3}}{2}t + C)\). Plugging into \(r(\theta) = \exp(-k\theta)\) and applying the boundary condition \(r(\theta_0) = \frac{\sqrt{3}}{3}\), we get \(r(t) = -\frac{\sqrt{3}}{2}t + \frac{\sqrt{3}}{3}\).
Next, \(\int_{\theta_0}^\infty r(\theta) d\theta = \frac{\sqrt{3}}{3k} = \frac{2}{3}\), so \(k=\frac{\sqrt{3}}{2}\) and \(\theta(t) = -\frac{2\sqrt{3}}{3} \ln(-\frac{\sqrt{3}}{2}t + \frac{\sqrt{3}}{3})\).