Questions: Question 20, 9.4.53 Section 9.4 Part 2 of 2 Graph the following system of inequalities. Find the coordinates of any vertices formed - 9x + 8y ≤ 72 - x + 3y ≤ 9 - x ≥ 0 - y ≥ 0 Use the graphing tool to graph the system. The coordinates of the vertex/vertices is/are (Type an ordered pair. Use a comma to separate answers as needed.)

Solution Steps

The given system of inequalities is: \[ \begin{array}{l} 9x + 8y \leq 72 \\ x + 3y \leq 9 \\ x \geq 0 \\ y \geq 0 \end{array} \]

To find the vertices, we need to solve the system of equations formed by the boundaries of the inequalities.

1. Solve \(9x + 8y = 72\) and \(x + 3y = 9\): \[ \begin{cases} 9x + 8y = 72 \\ x + 3y = 9 \end{cases} \]

Multiply the second equation by 9: \[ \begin{cases} 9x + 8y = 72 \\ 9x + 27y = 81 \end{cases} \]

Subtract the first equation from the second: \[ 19y = 9 \implies y = \frac{9}{19} \approx 0.4737 \]

Substitute \(y\) back into \(x + 3y = 9\): \[ x + 3\left(\frac{9}{19}\right) = 9 \implies x + \frac{27}{19} = 9 \implies x = 9 - \frac{27}{19} = \frac{144}{19} \approx 7.5789 \]

So, one vertex is \(\left(\frac{144}{19}, \frac{9}{19}\right) \approx (7.5789, 0.4737)\).

2. Solve \(9x + 8y = 72\) and \(x = 0\): \[ \begin{cases} 9(0) + 8y = 72 \\ x = 0 \end{cases} \] \[ 8y = 72 \implies y = 9 \]

So, another vertex is \((0, 9)\).

3. Solve \(x + 3y = 9\) and \(y = 0\): \[ \begin{cases} x + 3(0) = 9 \\ y = 0 \end{cases} \] \[ x = 9 \]

So, another vertex is \((9, 0)\).

The vertices of the feasible region are: \[ (0, 9), (9, 0), \left(\frac{144}{19}, \frac{9}{19}\right) \approx (7.5789, 0.4737) \]

Final Answer