= Solution
{c}
After removing the two redundant inequalities, the <feasible region> is
$$
y_1\ge0,\qquad
y_2\ge\max\left\{0,\ y_1-1,\ 3-\frac34y_1,\ 6-3y_1\right\}.
$$
Its lower boundary consists of three <line segments> or rays: $y_2=6-3y_1$ for $0\le y_1\le4/3$, then $y_2=3-3y_1/4$ for $4/3\le y_1\le16/7$, then $y_2=y_1-1$ for $y_1\ge16/7$. The <feasible region> is everything on or above this boundary in the first quadrant.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-oxford/2015/prelims/mathematics-iv-q5-dual.png]
{title=Dual feasible region and optimal supporting line in question 5}
The boundary <extreme points> are
$$
(0,6),\qquad (4/3,2),\qquad (16/7,9/7).
$$
Along its three successive pieces, the <objective function> $2y_1+y_2$ has respective slopes $-1$, $5/4$ and $3$ as $y_1$ increases. Thus it decreases to the middle <extreme point> and increases thereafter. For fixed $y_1$, moving above the boundary only increases the <objective function>. Hence
$$
\boxed{\mathbf y^*=(4/3,2),\qquad \min(2y_1+y_2)=\frac{14}{3}.}
$$
As an algebraic check of the graphical answer, the active inequalities give
$$
2y_1+y_2=\frac19(3y_1+4y_2)+\frac59(3y_1+y_2)
\ge\frac{12}{9}+\frac{30}{9}=\frac{14}{3},
$$
with equality at the stated <extreme point>.
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