= Solution
The <point> $(1,4)$ satisfies <dual feasibility> and supplies an upper bound of $13$ by <weak duality>. Directly, take one copy of the first constraint and four copies of the second:
$$
2x+3y+4z+9w\leq13.
$$
Since $w\geq0$, the original <objective function> is at most $13-2w\leq13$. The <feasible point> $(2,3,0,0)$ attains $13$. Hence
$$
\boxed{\operatorname{OPT}=13,\qquad (x,y,z,w)=(2,3,0,0)\text{ is optimal}.}
$$
There are also nonvertex optima. In terms of the <slack variables>, the exact <objective function> identity is
$$
2x+3y+4z+7w=13-s_1-4s_2-2w.
$$
Equality requires $s_1=s_2=w=0$, leaving the entire <optimal face>
$$
(x,y,z,w)=(2-t/2,\,3-t,\,t,\,0),\qquad t\geq0.
$$
Negative $x$ or $y$ on this ray are allowed. This illustrates why the single <vertex of a linear polyhedron> does not imply a unique optimum.
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