<p>The "golden rule" is the level at which steady-state consumption is at a maximum, given the parameters of the model. Steady state consumption is</p>

<p>$$c^* = (1-s^*)\cdot f[k^*(s^*)] = f[k^*(s^*)]  - s^*f[k^*(s^*)] \tag{1}$$</p>

<p>where $0&lt;s^*&lt;1$</p>

<p>We also have that, at the steady state (for constant capital)</p>

<p>$$s^*f[k^*(s^*)] = \delta k^*(s^*) \tag{2}$$</p>

<p>Inserting $(2)$ in $(1)$,</p>

<p>$$c^* =  f[k^*(s^*)]  - \delta k^*(s^*) \tag{3}$$</p>

<p>We want to maximize steady-state consumption, so we take the first derivative and set it equal to zero,</p>

<p>$$\frac {\partial c^*}{\partial s^*} = f'[k^*(s^*)]\cdot \frac {{\rm d}k^*}{{\rm d}s^*} - \delta \frac {{\rm d}k^*}{{\rm d}s^*} =0 $$</p>

<p>$$\implies \big(f'[k^*(s^*)]\ - \delta \big)\frac {{\rm d}k^*}{{\rm d}s^*} =0 \implies f'[k^*(s^*)] = \delta \tag{4} $$</p>

<p>From the production function we have</p>

<p>$$f'(k) = \frac 1{\sqrt {k}} = \delta \implies k^* = \frac 1 {\delta^2}$$</p>

<p>The rest are evident.</p>

CC BY-SA 3.0
2015-03-12T16:05:59.770Z
2015-03-12T16:05:59.770Z
Alecos Papadopoulos
<p>Since the depreciation rate is already a "derivative" (of the line $$\delta k$$), I'm not sure what you mean by "taking the derivative of the rate." If you take the derivative of the rate, you get zero.
In any case, if I understand your question, it's true you want to take the derivative of f, and set it equal to the depreciation rate, and solve for k.  What do you get when you take the derivate of f? What is this "simple power rule" you've used?</p>

CC BY-SA 3.0
2015-03-11T18:10:02.390Z
2015-03-11T18:10:02.390Z
user164740
<p>Consider an economy described by the per worker production function function:
 $$y = f(k) = 2k^\frac{1}{2}$$   and a depreciation rate $δ$ of $.05, 5%$.</p>

<p>Considering what we know to be true of the golden rule level of capital, find the golden rule level of capital $k_{gold}$ and output $y_{gold}$ using the above information.</p>

<p>Suppose there is no population growth rate $n = 0$.  Given the Solow model equation $(\Delta k = sf(k) – \delta k)$, and the answers for $k^*$ and $y^*$ you got in part A, what savings rate must this country have to put us at the golden rule level of capital?</p>

<p>Here is what I did:</p>

<blockquote>
  <p>In part A, I took the derivatives of the depreciation rate, and the
  per worker production function, and got that $K =.05$ just by the
  simple power rule. To find the output I plugged $K=.05$ back into the
  per worker production function to get $2(.05)^{1/2}$ and came up with
  an output of $.447$</p>
  
  <p>Part B: Since the equation I need now is $sf(k) = \delta k$ which
  using what I know, $s\times.447 =.05 \times .05$ Solving for $s$ I get
  that the savings rate is $0.556$ %.</p>
</blockquote>

<p>However, this is not correct. Please help me find the correct solution method and correct solution.</p>

<macroeconomics><economic-growth>
Deriving the golden-rule savings rate in a Solow Model
All About Groups
# Deriving the golden-rule savings rate in a Solow Model

(Asked by: username_0 on 2015-03-09)

Consider an economy described by the per worker production function function:
$$y = f(k) = 2k^\frac{1}{2}$$ and a depreciation rate $δ$ of $.05, 5%$.

Considering what we know to be true of the golden rule level of capital, find the golden rule level of capital $k\_{gold}$ and output $y\_{gold}$ using the above information.

Suppose there is no population growth rate $n = 0$. Given the Solow model equation $(\Delta k = sf(k) – \delta k)$, and the answers for $k^\*$ and $y^\*$ you got in part A, what savings rate must this country have to put us at the golden rule level of capital?

Here is what I did:

> In part A, I took the derivatives of the depreciation rate, and the
> per worker production function, and got that $K =.05$ just by the
> simple power rule. To find the output I plugged $K=.05$ back into the
> per worker production function to get $2(.05)^{1/2}$ and came up with
> an output of $.447$
>
> Part B: Since the equation I need now is $sf(k) = \delta k$ which
> using what I know, $s\times.447 =.05 \times .05$ Solving for $s$ I get
> that the savings rate is $0.556$ %.

However, this is not correct. Please help me find the correct solution method and correct solution.

**Tags:** macroeconomics, economic-growth

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## 2 answers

### Answer by username_1 on 2015-03-12. Score: 7

The "golden rule" is the level at which steady-state consumption is at a maximum, given the parameters of the model. Steady state consumption is

$$c^\* = (1-s^\*)\cdot f[k^\*(s^\*)] = f[k^\*(s^\*)] - s^\*f[k^\*(s^\*)] \tag{1}$$

where $0

We also have that, at the steady state (for constant capital)

$$s^\*f[k^\*(s^\*)] = \delta k^\*(s^\*) \tag{2}$$

Inserting $(2)$ in $(1)$,

$$c^\* = f[k^\*(s^\*)] - \delta k^\*(s^\*) \tag{3}$$

We want to maximize steady-state consumption, so we take the first derivative and set it equal to zero,

$$\frac {\partial c^\*}{\partial s^\*} = f'[k^\*(s^\*)]\cdot \frac {{\rm d}k^\*}{{\rm d}s^\*} - \delta \frac {{\rm d}k^\*}{{\rm d}s^\*} =0 $$

$$\implies \big(f'[k^\*(s^\*)]\ - \delta \big)\frac {{\rm d}k^\*}{{\rm d}s^\*} =0 \implies f'[k^\*(s^\*)] = \delta \tag{4} $$

From the production function we have

$$f'(k) = \frac 1{\sqrt {k}} = \delta \implies k^\* = \frac 1 {\delta^2}$$

The rest are evident.


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### Answer by username_2 on 2015-03-11. Score: 0

Since the depreciation rate is already a "derivative" (of the line $$\delta k$$), I'm not sure what you mean by "taking the derivative of the rate." If you take the derivative of the rate, you get zero.
In any case, if I understand your question, it's true you want to take the derivative of f, and set it equal to the depreciation rate, and solve for k. What do you get when you take the derivate of f? What is this "simple power rule" you've used?


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