Leontief function marginal product of labor/capital Find marginal product of labor of a Leontief production function.  For example,
$$f(L, K) = min\{\frac{L}{a}, \frac{K}{b}\}$$

MY ATTEMPT

Now as I understand it the marginal product of labor, $MP_L$ of this function cannot be found in the "normal" way because this function is not differentiable.  However, by definition, marginal product of labor means the extra output when one puts in an additional unit of labor.  We assume capital stays constant.

Say, we produce originally $q$ units of output.

Thus, if $\frac{L}{a}+\Delta L &lt; \frac{K}{b}$ where we are increasing the quantity of labor, then we are on the vertical portion of the isoquant (Labor on the horizontal axis, Capital on vertical).  Thus, we would increase our output by $q+\Delta L$.

If $\frac{L}{a}+\Delta L&gt;\frac{K}{b}$, we would still be producing $q$ units.
Since you are interested in labour, let's assume for simplicity that the stock of capital is fixed at $\bar{K}$. Then, the optimal choice of capital and labour is given by:

$$\frac{L^*}{a}=\frac{\bar{K}}{b}$$

Therefore, optimal labour is:

$$L^* = \frac{a}{b}\bar{K}$$

The marginal product of labour depends on how actual labour relates to optimal labour:


Case 1: $L = L^*$. In the standard Leontief diagram, with $L$ in the horizontal axis and $K$ in vertical axis, this is any point on the optimal path (which function starts at the origin and has slope $\frac{b}{a}$). In this case, $\dfrac{dQ}{dL}=0$.
Case 2: $L &gt; L^*$. This is when the factors' combination is below the $\frac{b}{a}$ path. In this case, $\dfrac{dQ}{dL}=0$.
Case 3: $L &lt; L^*$. This is when the factors' combination is above the $\frac{b}{a}$ path. The solution here depends on how far $L$ is from $L^*$:


Case 3a: $L^*-L &gt; 1$. This is perhaps the most likely case. Here, the change in labour leaves still with low levels of labour. In this scenario, $MP_L = \dfrac{1}{a}$. This results comes from comparing output before the change in labour: $Q_0=\dfrac{L_0}{a}$ versus after the change: $Q_1=\dfrac{L_0 +\Delta L}{a}$. From here, we conclude that $\dfrac{dQ}{dL}=\dfrac{1}{a}$. Notice that in this case, there is still room for increasing output by increasing labour (i.e. we are still within Case 3).
Case 3b: $L^*-L = 1$. Here, the change in labour leaves us on the optimal path. The change is just as in Case 3a. The differences is that increasing labour further leaves without increases in output. We are then back to case 1.
Case 3c: $L^*-L &lt; 1$. Here, the change in labour is more than what we actually require. Thus, we move from being in Case 3 to being in Case 2. The change is therefore not $\dfrac{1}{a}$, but equal to $\dfrac{1}{a} \times (L^*-L)$. In other words, the change is proportional to the exact amount of labour input we need to be in the optimal path.





A graphical example can be seen below:



For any labour equal or above $L_0$, further increases in $L$ does not change $q_0$. However, when we are above the optimal path (meaning $K&gt;\bar{K}$), the extra marginal worker does add to production. You can see that depending on how far this point is from the optimal, the three scenarios of Case 3 arise.

If interested (in my personal quest in favour of Open Economics), here is the R code of the graph:

plot(c(0,5), c(0,5), type = "n", xlab = "Labour", ylab = "Capital", yaxt='n', xaxt='n', bty='l', mgp=c(1,1,0))
segments(0, 0, 4.7, 4.7,lwd=2)
text(4.9, 4.8, expression(frac(b,a)),cex = 1.3)
points(2, 2, type="p", pch=19, col="black", bg=NA, cex=1.5)
segments(2, 2, 2, 5,lwd=2.5)
segments(2, 2, 4.7, 2,lwd=2.5)
text(4.9, 2,  expression(q[0]),cex = 1.3)
text(2.4, 1.75, expression(paste("(",L[0],",",bar(K),")")),cex = 1.3)
points(3.5, 2, type="p", pch=19, col="black", bg=NA, cex=1.5)
text(3.5, 2.3, expression(L&gt;L[0]),cex = 1.3)
points(2, 3.5, type="p", pch=19, col="black", bg=NA, cex=1.5)
text(2.4, 3.5, expression(K&gt;bar(K)),cex = 1.3)