Solve the Following System of Inequalities Graphically
`3x + 4y le 60, x ge 2y , x ge1 , y ge 0`

Explanation

To solve the system of inequalities graphically, we will follow these steps:

### Step 1: Convert Inequalities to Equations
We start with the inequalities:
1. \(3x + 4y \leq 60\)
2. \(x \geq 2y\)
3. \(x \geq 1\)
4. \(y \geq 0\)

We will first convert these inequalities into equations to find the boundary lines.

### Step 2: Graph the First Inequality
For the first inequality \(3x + 4y = 60\):
- Set \(x = 0\):
\[ 3(0) + 4y = 60 \implies 4y = 60 \implies y = 15 \quad \text{(Point: (0, 15))} \]
- Set \(y = 0\):
\[ 3x + 4(0) = 60 \implies 3x = 60 \implies x = 20 \quad \text{(Point: (20, 0))} \]

Now we plot the points (0, 15) and (20, 0) and draw the line \(3x + 4y = 60\). Since the inequality is less than or equal to, we shade below this line.

### Step 3: Graph the Second Inequality
For the second inequality \(x \geq 2y\):
- Set \(x = 0\):
\[ 0 \geq 2y \implies y \leq 0 \quad \text{(Point: (0, 0))} \]
- Set \(y = 0\):
\[ x \geq 2(0) \implies x \geq 0 \quad \text{(Point: (0, 0))} \]
- Set \(y = 2\):
\[ x = 2(2) = 4 \quad \text{(Point: (4, 2))} \]

Now we plot the points (0, 0) and (4, 2) and draw the line \(x = 2y\). Since the inequality is greater than or equal to, we shade above this line.

### Step 4: Graph the Third Inequality
For the third inequality \(x \geq 1\):
- This is a vertical line at \(x = 1\). We shade to the right of this line.

### Step 5: Graph the Fourth Inequality
For the fourth inequality \(y \geq 0\):
- This is the x-axis. We shade above this line.

### Step 6: Identify the Feasible Region
The feasible region is where all shaded areas overlap. We need to find the intersection points of the lines to determine the vertices of the feasible region.

### Step 7: Find Intersection Points
1. **Intersection of \(3x + 4y = 60\) and \(x = 2y\)**:
\[ 3(2y) + 4y = 60 \implies 6y + 4y = 60 \implies 10y = 60 \implies y = 6 \implies x = 2(6) = 12 \quad \text{(Point: (12, 6))} \]

2. **Intersection of \(3x + 4y = 60\) and \(x = 1\)**:
\[ 3(1) + 4y = 60 \implies 3 + 4y = 60 \implies 4y = 57 \implies y = \frac{57}{4} = 14.25 \quad \text{(Point: (1, 14.25))} \]

3. **Intersection of \(x = 2y\) and \(x = 1\)**:
\[ 1 = 2y \implies y = \frac{1}{2} \quad \text{(Point: (1, 0.5))} \]

4. **Intersection of \(x = 1\) and \(y = 0\)**:
\[ \text{Point: (1, 0)} \]

### Step 8: Plot the Feasible Region
Now we plot all the intersection points and shade the feasible region. The vertices of the feasible region are:
- (1, 0)
- (1, 14.25)
- (12, 6)
- (20, 0)

### Final Solution
The solution to the system of inequalities is the set of all points in the feasible region formed by these vertices.

---