Solve the following linear programming problem graphically:
Maximize : `z=30x+25y`
Subject to: `x+yle6`
`3x+2yle15`
`xge0`
`yge0`

Explanation

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

### Step 1: Define the objective function and constraints
We need to maximize the objective function:
\[ z = 30x + 25y \]

Subject to the constraints:
1. \( x + y \leq 6 \)
2. \( 3x + 2y \leq 15 \)
3. \( x \geq 0 \)
4. \( y \geq 0 \)

### Step 2: Graph the constraints
1. **Graph the first constraint** \( x + y = 6 \):
- When \( x = 0 \), \( y = 6 \) (point A: (0, 6))
- When \( y = 0 \), \( x = 6 \) (point B: (6, 0))
- Draw the line connecting points A and B, shading the area below the line since we have \( x + y \leq 6 \).

2. **Graph the second constraint** \( 3x + 2y = 15 \):
- When \( x = 0 \), \( y = 7.5 \) (point C: (0, 7.5))
- When \( y = 0 \), \( x = 5 \) (point D: (5, 0))
- Draw the line connecting points C and D, shading the area below the line since we have \( 3x + 2y \leq 15 \).

3. **Consider the non-negativity constraints** \( x \geq 0 \) and \( y \geq 0 \):
- This restricts our feasible region to the first quadrant.

### Step 3: Identify the feasible region
The feasible region is the area where the shaded regions from both constraints overlap, bounded by the axes.

### Step 4: Find the corner points of the feasible region
The corner points of the feasible region can be determined by finding the intersection points of the constraint lines and the axes:
1. Intersection of \( x + y = 6 \) and \( 3x + 2y = 15 \):
- Solve the equations simultaneously:
- From \( x + y = 6 \), we can express \( y = 6 - x \).
- Substitute into \( 3x + 2(6 - x) = 15 \):
\[ 3x + 12 - 2x = 15 \]
\[ x = 3 \]
\[ y = 6 - 3 = 3 \]
- Intersection point: (3, 3)

2. The other corner points are:
- Point A: (0, 6)
- Point D: (5, 0)

### Step 5: Evaluate the objective function at each corner point
Now we evaluate \( z \) at each of the corner points:
1. At point (0, 6):
\[ z = 30(0) + 25(6) = 150 \]

2. At point (3, 3):
\[ z = 30(3) + 25(3) = 90 + 75 = 165 \]

3. At point (5, 0):
\[ z = 30(5) + 25(0) = 150 \]

### Step 6: Determine the maximum value
Comparing the values of \( z \):
- At (0, 6): \( z = 150 \)
- At (3, 3): \( z = 165 \)
- At (5, 0): \( z = 150 \)

The maximum value of \( z \) is **165** at the point **(3, 3)**.

### Final Answer
The maximum value of \( z \) is **165** at the point \( (x, y) = (3, 3) \).

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