Questions: Graph the solution set to the follow system of inequalities by drawing a polygon around the solution region. Click to set the corner points. 2x+5y ≤ 40 6x+5y ≤ 60 x ≥ 0 y ≥ 0

Graph the solution set to the follow system of inequalities by drawing a polygon around the solution region. Click to set the corner points.

2x+5y ≤ 40
6x+5y ≤ 60
x ≥ 0
y ≥ 0

Transcript text: Graph the solution set to the follow system of inequalities by drawing a polygon around the solution region. Click to set the corner points. \[ \left\{\begin{array}{ll} 2 x+5 y & \leq 40 \\ 6 x+5 y & \leq 60 \\ x & \geq 0 \\ y & \geq 0 \end{array}\right. \] Clear All Draw: Polygon Question Help: Video Submit Question

Solution

Solution Steps

Step 1: Graph the Inequality \(2x + 5y \leq 40\)

To graph \(2x + 5y \leq 40\), first find the intercepts.
- When \(x = 0\): \(5y = 40 \Rightarrow y = 8\)
- When \(y = 0\): \(2x = 40 \Rightarrow x = 20\)
Plot the points \((0, 8)\) and \((20, 0)\) and draw the line.
Shade the region below the line since it is \(\leq\).

Step 2: Graph the Inequality \(6x + 5y \leq 60\)

To graph \(6x + 5y \leq 60\), first find the intercepts.
- When \(x = 0\): \(5y = 60 \Rightarrow y = 12\)
- When \(y = 0\): \(6x = 60 \Rightarrow x = 10\)
Plot the points \((0, 12)\) and \((10, 0)\) and draw the line.
Shade the region below the line since it is \(\leq\).

Step 3: Graph the Inequalities \(x \geq 0\) and \(y \geq 0\)

These inequalities represent the first quadrant of the coordinate plane.
Shade the region to the right of the \(y\)-axis and above the \(x\)-axis.

Step 4: Identify the Feasible Region

The feasible region is the intersection of all shaded regions from the previous steps.
This region is bounded by the lines \(2x + 5y = 40\), \(6x + 5y = 60\), \(x = 0\), and \(y = 0\).

Step 5: Find the Corner Points

The corner points are the intersections of the boundary lines.
- Intersection of \(2x + 5y = 40\) and \(6x + 5y = 60\):
  - Solve the system: \[ \begin{cases} 2x + 5y = 40 \\ 6x + 5y = 60 \end{cases} \]
  - Subtract the first equation from the second: \[ 4x = 20 \Rightarrow x = 5 \]
  - Substitute \(x = 5\) into \(2x + 5y = 40\): \[ 2(5) + 5y = 40 \Rightarrow 10 + 5y = 40 \Rightarrow 5y = 30 \Rightarrow y = 6 \]
  - Intersection point: \((5, 6)\)
- Intersection of \(2x + 5y = 40\) and \(x = 0\):
  - \(2(0) + 5y = 40 \Rightarrow y = 8\)
  - Intersection point: \((0, 8)\)
- Intersection of \(6x + 5y = 60\) and \(x = 0\):
  - \(6(0) + 5y = 60 \Rightarrow y = 12\)
  - Intersection point: \((0, 12)\)
- Intersection of \(x = 0\) and \(y = 0\):
  - Intersection point: \((0, 0)\)