Maximizing entropy in QM (Sakurai's Modern Quantum Mechanics)

There is section in Sakurai's "Modern Quantum Mechanics 2nd edition" page 188 that is quite confusing as to what he is doing. In the section on "Quantum Statistical Mechanics" he defines a quantity $\sigma = -\text{tr}(\rho \text{ln}\rho)$, where the entropy is given as $S = k \sigma$. Then in order to maximize $\sigma$ he states the following "Let us maximize $\sigma$ by requiring that $$\delta \sigma = 0.~~~~~~~~~~~~~(1)$$

However, we must take into account the constraint that the ensemble average of $H$ has a certain prescribed value. In the language of statistical mechanics, $[H]$ is identified with the internal energy per constituent, denoted by $U$: $$[H] = \text{tr}(\rho H) = U.$$ In addition, we should not forget the normalization constraint $\sum_{k} \rho_{kk} = 1$ (where $\rho$ is the density operator). So our basic task is to require (1) subject to the constraints $$\delta(H) = \sum_{k} \delta \rho_{kk} E_k = 0$$ and $$\delta(\text{tr} \rho) = \sum_{k} \delta \rho_{kk} = 0.$$

We can most readily accomplish this by using Lagrange multipliers. We obtain $$\sum_{k} \delta \rho_{kk}[(\text{ln}\rho_{kk} +1) + \beta E_{k} + \gamma] = 0$$ which for an arbitrary variation is possible only if $$\rho_{kk} = \text{exp}(- \beta E_{k} - \gamma - 1).$$ "

Question: What is the reasoning for the requirement (1) in order to maximize $\sigma$, and how are the constraints obtained?