= Solution
The point $(x_1,x_2,x_3)=(4,2,2)$ is nonnegative and makes all three inequalities equalities, with objective value $20$. To certify global optimality, combine the first upper bound with coefficient $1/2$, the second with coefficient one, and the lower bound with coefficient $-3/2$:
$$
\begin{aligned}
3x_1+2x_2+2x_3&=\tfrac12(7x_1+3x_2+5x_3)+(x_1+2x_2+x_3)-\tfrac32(x_1+x_2+x_3)\\
&\le\tfrac12\cdot44+10-\tfrac32\cdot8=20.
\end{aligned}
$$
This is a <weak duality> certificate for the <linear program>. The feasible point attains the bound, so \b[the optimal solution is $(4,2,2)$ and the maximum is $20$]. Equality forces each of the three constraints to be active; their simultaneous equations have this unique solution.
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