Questions: Graph the feasible region for the following system of inequalities by drawing a polygon around the feasible region. Click to set the corner points. x+y ≤ 11, -x+y ≤ 3, y ≥ 4.

Graph the feasible region for the following system of inequalities by drawing a polygon around the feasible region. Click to set the corner points. x+y ≤ 11, -x+y ≤ 3, y ≥ 4.

Transcript text: Graph the feasible region for the following system of inequalities by drawing a polygon around the feasible region. Click to set the corner points. $\left\{\begin{array}{ll}x+y & \leq 11 \\ -x+y & \leq 3 \\ y & \geq 4\end{array}\right.$

Solution

Solution Steps

Step 1: Graph the Inequality $x + y \leq 11$

To graph $x + y \leq 11$, first graph the line $x + y = 11$.
The intercepts are $ (11, 0) $ and $ (0, 11) $.
Shade the region below the line since $y$ is less than or equal to $11 - x$.

Step 2: Graph the Inequality $-x + y \leq 3$

To graph $-x + y \leq 3$, first graph the line $-x + y = 3$.
The intercepts are $ (-3, 0) $ and $ (0, 3) $.
Shade the region below the line since $y$ is less than or equal to $x + 3$.

Step 3: Graph the Inequality $y \geq 4$

To graph $y \geq 4$, draw a horizontal line at $y = 4$.
Shade the region above the line since $y$ is greater than or equal to 4.

Step 4: Identify the Feasible Region

The feasible region is the intersection of the shaded regions from the three inequalities.
Identify the corner points of the feasible region by finding the intersections of the lines:
- Intersection of $x + y = 11$ and $y = 4$: $x + 4 = 11 \Rightarrow x = 7$, so the point is $(7, 4)$.
- Intersection of $-x + y = 3$ and $y = 4$: $-x + 4 = 3 \Rightarrow x = 1$, so the point is $(1, 4)$.
- Intersection of $x + y = 11$ and $-x + y = 3$: Solve the system: \[ \begin{cases} x + y = 11 \\ -x + y = 3 \end{cases} \] Adding the equations: $2y = 14 \Rightarrow y = 7$, then $x + 7 = 11 \Rightarrow x = 4$, so the point is $(4, 7)$.