Solve the following linear programming problem graphically:
Minimize : `z=200x+500y`
Subject to: `x+2yge10`
`3x+4yle24`
`xge0`
`yge0`

Explanation

To solve the linear programming problem graphically, we will follow these steps:

### Step 1: Define the objective function and constraints
We need to minimize the function:
\[ z = 200x + 500y \]

Subject to the constraints:
1. \( x + 2y \geq 10 \)
2. \( 3x + 4y \leq 24 \)
3. \( x \geq 0 \)
4. \( y \geq 0 \)

### Step 2: Convert inequalities to equations
To graph the constraints, we first convert the inequalities into equations.

1. For \( x + 2y = 10 \):
- If \( x = 0 \), then \( 2y = 10 \) → \( y = 5 \) (Point: \( (0, 5) \))
- If \( y = 0 \), then \( x = 10 \) (Point: \( (10, 0) \))

2. For \( 3x + 4y = 24 \):
- If \( x = 0 \), then \( 4y = 24 \) → \( y = 6 \) (Point: \( (0, 6) \))
- If \( y = 0 \), then \( 3x = 24 \) → \( x = 8 \) (Point: \( (8, 0) \))

### Step 3: Graph the constraints
Now we will graph the lines on the coordinate plane.

- The line for \( x + 2y = 10 \) passes through points \( (0, 5) \) and \( (10, 0) \).
- The line for \( 3x + 4y = 24 \) passes through points \( (0, 6) \) and \( (8, 0) \).

### Step 4: Determine the feasible region
The feasible region is determined by the inequalities:
- For \( x + 2y \geq 10 \), we shade the area above the line.
- For \( 3x + 4y \leq 24 \), we shade the area below the line.

The feasible region is where these shaded areas overlap, and it must also be in the first quadrant (where \( x \geq 0 \) and \( y \geq 0 \)).

### Step 5: Find the corner points of the feasible region
The corner points of the feasible region can be found by solving the equations of the lines:
1. Set \( x + 2y = 10 \) and \( 3x + 4y = 24 \) together:
- From \( x + 2y = 10 \), express \( x \) in terms of \( y \):
\[ x = 10 - 2y \]
- Substitute into \( 3x + 4y = 24 \):
\[ 3(10 - 2y) + 4y = 24 \]
\[ 30 - 6y + 4y = 24 \]
\[ -2y = -6 \]
\[ y = 3 \]
- Substitute \( y = 3 \) back to find \( x \):
\[ x = 10 - 2(3) = 4 \]
- So, one corner point is \( (4, 3) \).

2. The other corner points are:
- \( (0, 5) \) from the first constraint.
- \( (0, 6) \) from the second constraint.
- \( (8, 0) \) from the second constraint.

### Step 6: Evaluate the objective function at each corner point
Now we evaluate \( z = 200x + 500y \) at each corner point:
1. At \( (4, 3) \):
\[ z = 200(4) + 500(3) = 800 + 1500 = 2300 \]
2. At \( (0, 5) \):
\[ z = 200(0) + 500(5) = 0 + 2500 = 2500 \]
3. At \( (0, 6) \):
\[ z = 200(0) + 500(6) = 0 + 3000 = 3000 \]
4. At \( (8, 0) \):
\[ z = 200(8) + 500(0) = 1600 + 0 = 1600 \]

### Step 7: Identify the minimum value
The minimum value of \( z \) occurs at the point \( (4, 3) \) with \( z = 2300 \).

### Final Answer
The minimum value of \( z \) is \( 2300 \) at the point \( (4, 3) \).

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