Deriving the golden-rule savings rate in a Solow Model 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.
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 $0&lt;s^*&lt;1$

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.