Deriving the golden-rule savings rate in a Solow Model

Question

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.

Answer 1

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 $0<s^*<1$

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.

Answer 2

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?