Q.Minimize $Z = 7x + y$ subject to $5x + y \geq 5$, $x + y \geq 3$, $x \geq 0$, $y \geq 0$.

Plot the feasible region for $5x+y\geq5$, $x+y\geq3$, $x,y\geq0$; evaluate $Z$ at each corner point.

Boundary lines:

$5x+y=5$ passes through $(1,0)$ and $(0,5)$.

$x+y=3$ passes through $(3,0)$ and $(0,3)$.

Feasible region: since both constraints are $\geq$, the region lies above both lines (in the first quadrant), and is unbounded.

Corner points:

• On the $y$-axis ($x=0$): need $y\geq5$ (from $5x+y\geq5$) and $y\geq3$ (from $x+y\geq3$); the more restrictive is $y\geq5$, giving vertex $(0,5)$.
• On the $x$-axis ($y=0$): need $x\geq1$ (from $5x\geq5$) and $x\geq3$ (from $x+y\geq3$); the more restrictive is $x\geq3$, giving vertex $(3,0)$.
• Intersection of the two lines: solving $5x+y=5$ and $x+y=3$ simultaneously — subtract to get $4x=2 \implies x=0.5$, then $y=2.5$. This gives vertex $(0.5, 2.5)$.

The feasible region (unbounded, extending away from the origin) has corner points $(0,5)$, $(0.5,2.5)$, and $(3,0)$.

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