Determine graphically the minimum value of the objective function: `Z = -50x + 20y` subject to the constraints:
`2x - y ge -5, 3x + y ge 3, 2x - 3y le 12, x, y ge 0`.

Answer

To solve the problem graphically, we need to follow these steps:

### Step 1: Write down the objective function and constraints
The objective function is given as:
\[ Z = -50x + 20y \]

The constraints are:
1. \( 2x - y \geq -5 \)
2. \( 3x + y \geq 3 \)
3. \( 2x - 3y \leq 12 \)
4. \( x, y \geq 0 \)

### Step 2: Convert inequalities into equations
We will convert each inequality into an equation to find the boundary lines.

1. For \( 2x - y = -5 \):
- Rearranging gives \( y = 2x + 5 \)

2. For \( 3x + y = 3 \):
- Rearranging gives \( y = -3x + 3 \)

3. For \( 2x - 3y = 12 \):
- Rearranging gives \( y = \frac{2}{3}x - 4 \)

### Step 3: Find intercepts for each line
To plot these lines, we can find the x and y intercepts.

1. For \( y = 2x + 5 \):
- x-intercept: Set \( y = 0 \) → \( 0 = 2x + 5 \) → \( x = -2.5 \) (not valid since \( x \geq 0 \))
- y-intercept: Set \( x = 0 \) → \( y = 5 \)

2. For \( y = -3x + 3 \):
- x-intercept: Set \( y = 0 \) → \( 0 = -3x + 3 \) → \( x = 1 \)
- y-intercept: Set \( x = 0 \) → \( y = 3 \)

3. For \( y = \frac{2}{3}x - 4 \):
- x-intercept: Set \( y = 0 \) → \( 0 = \frac{2}{3}x - 4 \) → \( x = 6 \)
- y-intercept: Set \( x = 0 \) → \( y = -4 \) (not valid since \( y \geq 0 \))

### Step 4: Plot the lines on a graph
- Plot the points (0, 5) for the first line.
- Plot the points (1, 0) and (0, 3) for the second line.
- Plot the point (6, 0) for the third line.

### Step 5: Determine the feasible region
- For the first constraint \( 2x - y \geq -5 \), shade the area above the line \( y = 2x + 5 \).
- For the second constraint \( 3x + y \geq 3 \), shade the area above the line \( y = -3x + 3 \).
- For the third constraint \( 2x - 3y \leq 12 \), shade the area below the line \( y = \frac{2}{3}x - 4 \).
- The feasible region is where all shaded areas overlap and is constrained to the first quadrant (where \( x, y \geq 0 \)).

### Step 6: Identify the corner points of the feasible region
The corner points of the feasible region can be found by solving the equations of the lines where they intersect.

1. Intersection of \( y = 2x + 5 \) and \( y = -3x + 3 \):
\[ 2x + 5 = -3x + 3 \implies 5x = -2 \implies x = -\frac{2}{5} \text{ (not valid)} \]

2. Intersection of \( y = -3x + 3 \) and \( y = \frac{2}{3}x - 4 \):
\[ -3x + 3 = \frac{2}{3}x - 4 \implies -\frac{11}{3}x = -7 \implies x = \frac{21}{11}, y = -3(\frac{21}{11}) + 3 = \frac{3}{11} \text{ (valid)} \]

3. Intersection of \( y = 2x + 5 \) and \( y = \frac{2}{3}x - 4 \):
\[ 2x + 5 = \frac{2}{3}x - 4 \implies \frac{4}{3}x = -9 \implies x = -\frac{27}{4} \text{ (not valid)} \]

4. The only valid corner point is \( (0, 5) \).

### Step 7: Evaluate the objective function at the corner points
Now we evaluate the objective function \( Z = -50x + 20y \) at the valid corner point \( (0, 5) \):
\[ Z = -50(0) + 20(5) = 100 \]

### Conclusion
The minimum value of the objective function \( Z \) is \( 100 \) at the point \( (0, 5) \).