Let $A$ = $\begin{pmatrix} 3&0&1\\ -1&2&-1\\ -2&-2&1 \end{pmatrix}$ and $X$ square matrix of order 3 diagonalizable and meets $AX=XA$. When $Tr(AX) = d$ , I need to find the maximum of $det(AX)$ as a function of $d$.
All eigenvalues of $X$ are positive.
What I have so far :
I know that $Tr(AX) = \sum_k{\lambda_k}$ and $Det(AX) = \prod_k{\lambda_k}$ with $\lambda_k$ eigenvalues of $AX$.
So to maximize $Det(AX)$ I can maximize $Ln(Det(AX)) = \sum_k{Ln(\lambda_k)}$.
Therefore I have $\dfrac{\partial Ln(Det(AX))}{\partial \lambda_k} = \dfrac{1}{\lambda_k}$. But I don't know how to continue.
$A$ has the diagonalisation $A=P\operatorname{diag}(1,2,3)P^{-1}$. Since $A$ possesses a real spectrum with distinct eigenvalues, every $X$ that commutes with $A$ must be in the form of $PDP^{-1}$ for some diagonal matrix $D=\operatorname{diag}(x,y,z)$. Therefore, your optimisation problem reduces to maximising $x(2y)(3z)$ subject to $x+2y+3z=d$ and $x,y,z>0$.
It follows that a feasible solution exists if and only if $d>0$. In this is the case, by the AM-GM inequality, the maximum clear occurs when $x=2y=3z=d/3$, i.e. when $(x,y,z)=(\frac d3,\frac d6,\frac d9)$ and the maximum value is given by $d^3/27$.