Past exam of the mathematics course of the University of Oxford / 2015 / Prelims / Mathematics IV / 5 / b / iii / Solution

After removing the two redundant inequalities, the feasible region isIts lower boundary consists of three line segments or rays: $y_2=6-3y_1$ for $0\le y_1\le 4/3$, then $y_2=3-3y_1/4$ for $4/3\le y_1\le 16/7$, then $y_2=y_1-1$ for $y_1\ge 16/7$. The feasible region is everything on or above this boundary in the first quadrant.

The boundary extreme points areAlong 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. HenceAs an algebraic check of the graphical answer, the active inequalities givewith equality at the stated extreme point.