Questions: Question 5 Suppose the production function is y=10k^0.3, the depreciation rate is 5%, the population growth rate is 1%, the rate of labor-augmenting technological progress is 3%, and the saving rate is 25%. What is the steady-state capital stock per effective worker? a 88 b 94 c 101 d 115

To find the steady-state capital stock per effective worker, we need to use the Solow growth model with technological progress. The steady-state condition for capital per effective worker \( k^* \) is given by:

\[ s f(k^_) = (\delta + n + g) k^_ \]

where:

• \( s \) is the saving rate,
• \( f(k) \) is the production function per effective worker,
• \( \delta \) is the depreciation rate,
• \( n \) is the population growth rate,
• \( g \) is the rate of labor-augmenting technological progress.

Given:

• Production function: \( y = 10 k^{0.3} \)
• Depreciation rate (\( \delta \)): 5% or 0.05
• Population growth rate (\( n \)): 1% or 0.01
• Rate of labor-augmenting technological progress (\( g \)): 3% or 0.03
• Saving rate (\( s \)): 25% or 0.25

First, we need to find the steady-state condition:

\[ s f(k^_) = (\delta + n + g) k^_ \]

Substitute the given values:

\[ 0.25 \cdot 10 (k^_)^{0.3} = (0.05 + 0.01 + 0.03) k^_ \]

\[ 2.5 (k^_)^{0.3} = 0.09 k^_ \]

To solve for \( k^* \), divide both sides by \( k^* \):

\[ 2.5 (k^*)^{-0.7} = 0.09 \]

\[ (k^*)^{-0.7} = \frac{0.09}{2.5} \]

\[ (k^*)^{-0.7} = 0.036 \]

Raise both sides to the power of \(-\frac{1}{0.7}\):

\[ k^* = (0.036)^{-\frac{1}{0.7}} \]

\[ k^* = (0.036)^{-1.4286} \]

\[ k^* \approx 101 \]

Therefore, the steady-state capital stock per effective worker is approximately 101.

The answer is C: 101.