# Golden rule level of accumulation in Solow model

I am having trouble with the following problem:

The steady-state level of capital with the highest consumption is called the golden rule level of accumulation and is denoted by $k_{gold}$. Assuming that the depreciation rate $\delta$ is fixed and the savings rate $s$ is variable, show that in a Solow model with Cobb-Douglas production function $f(k) = k^\alpha$, the golden rule level $k_{gold}$ is achieved when $s = \alpha$.

Here is what I have done:

We have the consumption rate, $c = (1 - s) k^\alpha$. Here, both $s$ and $k$ are variables so we assume that c is a function of $s$ and $k$. To maximize c, we will find its critical points by taking its first derivatives: $\frac{\partial c}{\partial s} = -k^\alpha$ and $\frac{\partial c}{\partial k} = \alpha k^{\alpha-1} - s\alpha k^{\alpha-1}$. But when k > 0, $-k^\alpha = 0$ is impossible.

Here is another try:

We have $c = (1 - s) k^\alpha = k^\alpha - sk^\alpha$. We also have the differential equation, $\frac{dk}{dt} = sk^\alpha - \delta k$. At the steady state, $sk^\alpha = \delta k$. Thus, we rewrite our equation as $c = k^\alpha - \delta k$. Now, c is a function of k, and we take the first derivative: $\frac{dc}{dk} = \alpha k^{\alpha - 1} - \delta$. By setting it to zero, we get $k = (\frac{\delta}{\alpha})^\frac1{\alpha - 1}$. However, this also fails to prove that $k_{gold}$ is achieved at $s = \alpha$.

I have got it. In my second approach, I got $k = (\frac{\delta}{\alpha})^\frac1{\alpha-1}$. It is the only critical point for $c(k) = k^\alpha - \delta k$.
$$k<(\frac{\delta}{\alpha})^\frac1{\alpha-1} \Leftrightarrow k^{\alpha-1} > \frac{\delta}{\alpha} (\because 0<\alpha<1) \Leftrightarrow \alpha k^{\alpha-1}-\delta>0 \Leftrightarrow \frac{dc}{dk} > 0$$
Similarly, we have $k>(\frac{\delta}{\alpha})^\frac1{\alpha-1} \Leftrightarrow \frac{dc}{dk} < 0$. Thus, it is indeed a maximum and the global maximum. It is the steady-state level of capital at which the consumption rate is maximized. We could rewrite it as $k_{gold} = (\frac{\alpha}{\delta})^\frac1{1-\alpha}$.
Now, considering $\frac{dk}{dt} = sk^\alpha - \delta k$, the non-zero equilibrium solution is $k_s = (\frac s{\delta})^\frac1{1-\alpha}$. Therefore, when $s = \alpha$, $k_s = k_{gold}$. In other words, $k_{gold}$ is achieved at $s = \alpha$.