Q.Solve graphically: $4x + 3y \le 12$. State the type of boundary line and identify the solution half-plane.

Plot the boundary lines of each inequality, shade the correct side for each, and identify the common (feasible) region as the triangle formed by the three lines' pairwise intersections.

The system is $4x+3y\le60$, $y\ge2x$, $x\ge3$, $x,y\ge0$.

Boundary lines:

• $4x+3y=60$ — solution region is on or below this line (since $\le$)
• $y=2x$ — solution region is on or above this line (since $\ge$)
• $x=3$ — solution region is on or to the right of this vertical line (since $\ge$)
• $x\ge0,y\ge0$ — first quadrant (automatically satisfied once $x\ge3$)

Finding the vertices of the feasible region by solving the boundary lines pairwise:

$x=3$ and $y=2x$: $y=6 \Rightarrow (3,6)$. Check $4(3)+3(6)=30\le60$ ✓

$x=3$ and $4x+3y=60$: $12+3y=60 \Rightarrow y=16 \Rightarrow (3,16)$. Check $16\ge2(3)=6$ ✓

$y=2x$ and $4x+3y=60$: $4x+6x=60 \Rightarrow x=6, y=12 \Rightarrow (6,12)$. Check $x=6\ge3$ ✓

All three constraints are satisfied (with equality on the relevant boundary) at each of these points, and the region enclosed between them (bounded on the left by $x=3$, above by $y=2x$, and on the upper-right by $4x+3y=60$) is the feasible region.

Graphically: the feasible region is the closed triangular region with vertices $(3,6)$, $(3,16)$ and $(6,12)$.