Question

Consider a Solow model with the production function: \(Y=K^\alpha L^{1-\alpha}\), where \(Y\), \(K\), and \(L\) are output, capital, and labour, respectively. \(\alpha\) is a positive constant. The savings rate, depreciation rate, and labour growth rates are \(20\%\), \(7\%\), and \(3\%\), respectively. If \(\alpha=0.5\), then the steady-state capital-labour ratio is (in integer).

Hint:

In the Solow model, steady state occurs when \(sy=(n+\delta)k\). For Cobb-Douglas production, first convert the production function into per-worker form.

Answer 1

Correct Answer: 4

Step 1: Write the production function in per-worker form.
The given production function is
\[ Y=K^\alpha L^{1-\alpha} \]
Divide both sides by \(L\):
\[ \frac{Y}{L}=\frac{K^\alpha L^{1-\alpha}}{L} \]
\[ y=\left(\frac{K}{L}\right)^\alpha \]
Let
\[ k=\frac{K}{L} \]
Therefore,
\[ y=k^\alpha \]

Step 2: Recall the steady-state condition in the Solow model.
At steady state, investment per worker equals the break-even investment.
So,
\[ sy=(n+\delta)k \]
where \(s\) is savings rate, \(n\) is labour growth rate, and \(\delta\) is depreciation rate.

Step 3: Substitute the production function.
Since
\[ y=k^\alpha \] we get
\[ sk^\alpha=(n+\delta)k \]

Step 4: Substitute the given values.
Given,
\[ s=20\%=0.20 \] \[ \delta=7\%=0.07 \] \[ n=3\%=0.03 \] and
\[ \alpha=0.5 \]
Thus,
\[ n+\delta=0.03+0.07=0.10 \]
Substitute in the steady-state equation:
\[ 0.20k^{0.5}=0.10k \]

Step 5: Solve for steady-state capital-labour ratio.
\[ \frac{0.20}{0.10}= \frac{k}{k^{0.5}} \]
\[ 2=k^{0.5} \]
Squaring both sides,
\[ k=4 \]

Step 6: Final conclusion.
Hence, the steady-state capital-labour ratio is
\[ \boxed{4} \]