Q.Solve the following system of inequalities graphically: $4x+3y \le 60,\ y \ge 2x,\ x \ge 3,\ x, y \ge 0$

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 $x=6\ge3$ ✓

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