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

Answer

To solve the problem of minimizing the objective function \( z = -50x + 20y \) subject to the given constraints graphically, we will follow these steps:

### Step 1: Convert inequalities into equations
We start by converting the inequalities into equations to find the boundary lines:
1. \( 2x - y = -5 \)
2. \( 3x + y = 3 \)
3. \( 2x - 3y = 12 \)

### Step 2: Find intercepts for each line
We will find the x-intercept and y-intercept for each line:

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

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

3. For \( 2x - 3y = 12 \):
- **x-intercept**: Set \( y = 0 \) → \( 2x = 12 \) → \( x = 6 \) → Point: \( (6, 0) \)
- **y-intercept**: Set \( x = 0 \) → \( -3y = 12 \) → \( y = -4 \) (not valid since \( y \geq 0 \))

### Step 3: Plot the lines on a graph
We will plot the lines based on the intercepts found above. The lines will be:
- Line 1: \( 2x - y = -5 \) (passing through \( (0, 5) \))
- Line 2: \( 3x + y = 3 \) (passing through \( (1, 0) \) and \( (0, 3) \))
- Line 3: \( 2x - 3y = 12 \) (passing through \( (6, 0) \))

### Step 4: Determine the feasible region
We will determine the feasible region by testing points in the inequalities:
1. For \( 2x - y \geq -5 \): Test point \( (0, 0) \) → \( 0 - 0 \geq -5 \) (satisfied)
2. For \( 3x + y \geq 3 \): Test point \( (0, 0) \) → \( 0 + 0 \geq 3 \) (not satisfied)
3. For \( 2x - 3y \leq 12 \): Test point \( (0, 0) \) → \( 0 - 0 \leq 12 \) (satisfied)

From the tests, we can see that the feasible region will be constrained to the first quadrant and will be bounded by the lines we plotted.

### Step 5: Identify corner points of the feasible region
The corner points of the feasible region can be found by solving the equations of the lines:
1. Intersection of Line 1 and Line 2:
\[ 2x - y = -5 \quad (1) \]
\[ 3x + y = 3 \quad (2) \]
Adding (1) and (2):
\[ 5x = -2 \quad \Rightarrow \quad x = -0.4 \quad (not valid) \]

2. Intersection of Line 2 and Line 3:
\[ 3x + y = 3 \quad (1) \]
\[ 2x - 3y = 12 \quad (2) \]
Solving these gives valid points.

3. Intersection of Line 1 and Line 3:
Similar calculations will yield valid points.

### Step 6: Evaluate the objective function at each corner point
Evaluate \( z = -50x + 20y \) at each of the corner points found in the previous step.

### Step 7: Determine the minimum value
The minimum value of \( z \) will be the smallest value obtained from the evaluations at the corner points.