Question

graph the solution set to the following system of inequalities and identify each vertex of the region.
x≥0, y≥0, x + y ≤4, 2x + y ≤6

use the graphing tool to graph the system.

what are the vertices of the region?
(type an ordered pair. use a comma to separate answers as needed.)

Explanation:

Step1: Find intersection with axes

• For \(x = 0\):
• From \(y\geq0\) and \(x + y\leq4\), when \(x = 0\), \(y\leq4\), so one vertex is \((0,0)\) (intersection of \(x = 0\) and \(y = 0\)) and \((0,4)\) (intersection of \(x = 0\), \(y\geq0\) and \(x + y\leq4\)). Also, check \(2x + y\leq6\) at \(x = 0\), \(y\leq6\), but \(y\leq4\) is more restrictive, so \((0,4)\) is valid.
• From \(2x + y\leq6\) and \(x = 0\), \(y\leq6\), but combined with \(x + y\leq4\), \(y\) is limited by \(x + y\leq4\) when \(x = 0\).
• For \(y = 0\):
• From \(x\geq0\) and \(x + y\leq4\), when \(y = 0\), \(x\leq4\). From \(2x + y\leq6\), when \(y = 0\), \(2x\leq6\Rightarrow x\leq3\). So the intersection of \(y = 0\), \(x\geq0\) and \(2x + y\leq6\) is \((3,0)\), and intersection of \(y = 0\), \(x\geq0\) and \(x + y\leq4\) is \((4,0)\), but \(2x + y\leq6\) is more restrictive, so \((3,0)\) is valid, and \((0,0)\) is also from \(x = 0,y = 0\).

Step2: Find intersection of \(x + y=4\) and \(2x + y = 6\)

• Subtract the first equation from the second: \((2x + y)-(x + y)=6 - 4\)
• Simplify: \(2x + y - x - y=2\Rightarrow x = 2\)
• Substitute \(x = 2\) into \(x + y=4\): \(2 + y=4\Rightarrow y = 2\). So the intersection point is \((2,2)\)

Step3: List all vertices

• Intersection of \(x = 0,y = 0\): \((0,0)\)
• Intersection of \(x = 0,x + y\leq4\) (and \(2x + y\leq6\)): \((0,4)\)
• Intersection of \(x + y = 4\) and \(2x + y = 6\): \((2,2)\)
• Intersection of \(y = 0,2x + y\leq6\) (and \(x\geq0\)): \((3,0)\)

Answer:

\((0,0),(0,4),(2,2),(3,0)\)