<p>The condition
<span class="math-container">$$
f(k^\ast) = \frac{n + g + \delta}{s} k^\ast.
$$</span>
is an equilibrium condition which <strong>only</strong> holds at the value of <span class="math-container">$k^\ast$</span>. As such, you are not allowed to take derivatives of both sides and equate them.</p>
<p>Taking derivatives would only be allowed if this equation holds for all <span class="math-container">$k$</span> in a neighbourhood of <span class="math-container">$k^\ast$</span> (which it doesn't).</p>

CC BY-SA 4.0
2023-11-13T06:52:32.333Z
2023-11-13T06:52:32.333Z
tdm
<p>According to <a href="https://www.csus.edu/indiv/v/vangaasbeckk/courses/200a/sup/growthrates.pdf" rel="nofollow noreferrer">this primer on the Solow Growth model</a>, wages are equal to <span class="math-container">$w = \frac{\partial Y}{\partial L} = A \left(f(k) - f^{\prime}(k)k\right)$</span>.</p>
<p>In the balanced growth path we know that <span class="math-container">$\dot{k} = 0 = sf(k^{\star}) - (n+g+\delta)k^{\star}$</span> where <span class="math-container">$k^{\star}$</span> is the value of <span class="math-container">$k$</span> on the balanced growth path.</p>
<p>From this it follows that</p>
<p><span class="math-container">$$
f(k^{\star}) = \frac{n+g+\delta}{s}k^{\star}
$$</span>
<span class="math-container">$$
\implies f^{\prime}(k^{\star}) = \frac{n+g+\delta}{s}
$$</span>
<span class="math-container">$$
\implies f(k^{\star}) - f^{\prime}(k^{\star})k^{\star} = 0
$$</span></p>
<p>And so wages are zero.</p>
<p>But the primer also says that the growth rate of wages is <span class="math-container">$\frac{\dot{w}}{w} = g$</span> on the balanced growth path which doesn't make if we are dividing be <span class="math-container">$w=0$</span>. So there must be a flaw in the above logic, but I don't see it.</p>

<wages><solow><balanced-growth>
In Solow model, on the balanced growth path, why aren't wages zero?
gb1729
# In Solow model, on the balanced growth path, why aren't wages zero?

(Asked by: username_0 on 2023-11-12)

According to [this primer on the Solow Growth model](https://www.csus.edu/indiv/v/vangaasbeckk/courses/200a/sup/growthrates.pdf), wages are equal to $w = \frac{\partial Y}{\partial L} = A \left(f(k) - f^{\prime}(k)k\right)$.

In the balanced growth path we know that $\dot{k} = 0 = sf(k^{\star}) - (n+g+\delta)k^{\star}$ where $k^{\star}$ is the value of $k$ on the balanced growth path.

From this it follows that

$$
f(k^{\star}) = \frac{n+g+\delta}{s}k^{\star}
$$
$$
\implies f^{\prime}(k^{\star}) = \frac{n+g+\delta}{s}
$$
$$
\implies f(k^{\star}) - f^{\prime}(k^{\star})k^{\star} = 0
$$

And so wages are zero.

But the primer also says that the growth rate of wages is $\frac{\dot{w}}{w} = g$ on the balanced growth path which doesn't make if we are dividing be $w=0$. So there must be a flaw in the above logic, but I don't see it.

**Tags:** wages, solow, balanced-growth


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## 1 answers

### Answer by username_1 on 2023-11-13. Score: 1 | Accepted answer

The condition
$$
f(k^\ast) = \frac{n + g + \delta}{s} k^\ast.
$$
is an equilibrium condition which **only** holds at the value of $k^\ast$. As such, you are not allowed to take derivatives of both sides and equate them.

Taking derivatives would only be allowed if this equation holds for all $k$ in a neighbourhood of $k^\ast$ (which it doesn't).


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