Let there be $p_1,p_2,..,p_n,q_1,q_2,...,q_n$ with $\sum_{i=1}^n p_i = 1 = \sum_{i=1}^nq_i$
For $M :=\{ x\in (0,\infty)^n: \sum_{i=1}^nq_ix_i = a \}$
With the following Maximization Problem.
$$(*) \sup \bigg\{\sum_{i=1}^np_i\ln (x_i) : x \in M \bigg \}$$
(i) I need to show: $\forall\eta >0$ $$\sup_{r>0}(\ln(r) -\eta r) = \ln(\frac{1}{\eta})-1$$
(ii) I also need to show: $\forall x\in M$ and $\forall \eta > 0$ $$\sum_{i=0}^np_i\ln(x_i) \leq \sum_{i=0}^np_i\ln(\frac{p_i}{\eta q_i}) + a\eta -1 $$
(iii) I need to also show, that for the problem $(*)$ there is a maximizer $x^*\in M$ and to determine it's value.
I currently have no idea how to approach this, as this is my first maximization problem. For now i know that for the maximizer i can use the Lagrange-Method to obtain the maximum of this function. But for the issues (i) and (ii) i got a bit more trouble figuring out what i need to do.
Any help is appreciated.
i) Take the first derivative of the function $f(r)=ln(r)-\eta r$ then do some calculation.
ii) Using i) with $r=x_i$ and $\eta=\dfrac{\eta q_i}{p_i}$ then multiple with $p_i$ and sum up all the inequality.
iii) Notice that in ii) equality happens when $x_i=\dfrac{p_i}{\eta q_i}$. We need $\displaystyle\sum_{i=1}^n q_i x_i=a$, which is equivalent to $\eta =\dfrac{1}{a}$. Therefore, by just taking $\eta =\dfrac{1}{a}$ in ii), we're done.