Verified
(a) Given, the production function is Y = K0.2 (EL)0.8
In order to prove that this indicates constant returns to scale, the output in the production function, Y has to increase by the same proportion, which is used to increase all the inputs. In our case, if K and L increase by m, the output Y has to increase by m.
Suppose L and K increases by m, the new production function will be
Y’ = (mK)0.2 (mEL)0.8 = m0.2+0.8 K0.2 (EL)0.8 = m K0.2 (EL)0.8 = m*Y
Hence the output also increases by m. Thus this production function indicates constant returns to scale.
(b) From the production function, Y = K0.2 (EL)0.8
The marginal product of labor can be derived as ΔY/ΔL = 0.8 K0.2 (EL)-0.2 = 0.8(K/EL)0.2
From this derived equation, as L increases, the marginal product of labor will fall (since L is in the denominator). As more workers are hired, the extra output obtained from each additional new worker will fall as L increases and marginal product of labor will fall. Thus the production function indicates a decreasing marginal product of labor.
(c) As we defined k = K/EL and y = Y/EL, and we include our production function into it,
y = Y/EL = (K0.2 (EL)0.8)/EL = K0.2/EL0.2 = (K/EL)0.2 = k0.2
y = k0.2
Thus y is expressed as a function of k
(d) Labor, L grows at the rate of n (population growth rate), efficiency of labor, E grows at the rate of g (growth rate of labor efficiency level and Capital stock, K is depreciating at the level of δ (depreciation rate of capital). Since k = K / L *E, we can see how k changes over time:
dk = dK/EL – (K/EL2) dL - (K/LE2) dE
dk = (K/EL) dK/K – (K/EL) dL/L – (K/EL) dE/E
dk = kδ – kn – kg
Here the sign of kδ is also negative, since capital is consumed by depreciation (dK/K < 0).
In the steady state condition, Δk = 0
We also know that Δk = s*f(k) – δk
In our case, Δk = s*f(k) – (δ+g+n)*k
Since Δk = 0, s*f(k) = (δ+g+n)*k
k*/f(k) = s/ (δ+g+n)
k/k0.2 = s/ (δ+g+n)
k0.8 = s/ (δ+g+n)
This is the steady state level for k. Since we already know y = k0.2 (from (c)), at steady state, y* = (k*)0.2
Thus y* and k* are determined.
(e) All details given for North and South, they are as such substituted in k* and y*.
kN0.8 = 0.48/(0.01+0.02+0) = 0.48/0.03 = 16
kN* = 32
yN* = 2
kS0.8 = 0.09/(0.01+0.02+0.06) = 0.09/0.09 = 1
kS* = 1
yS* = 1
The steady state level of y in the North and the South are 2 and 1 respectively.