= Solution
For any feasible primal vector, twice the second constraint plus one third of the third gives
$$
-\frac23x_1+4x_2+3x_3-\frac{17}3x_4\geq\frac{25}3.
$$
Since $x_1,x_4\geq0$,
$$
2x_1+4x_2+3x_3+x_4
=\left(-\frac23x_1+4x_2+3x_3-\frac{17}3x_4\right)
+\frac83x_1+\frac{20}3x_4\geq\frac{25}3.
$$
The vector $(0,11/6,1/3,0)$ is feasible and attains this bound. \b[It is optimal], by this direct inequality proof of <weak duality>, without assuming the <simplex method> or a duality theorem. Equality forces $x_1=x_4=0$ and both positively weighted constraints tight, so it also proves uniqueness.
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