Consider the following LPP:
Maximize `Z = 3x + 2y` subject to the constraints:
`x + 2y le 10, 3x + y le 15, x, y ge 0`.
(a) Draw the feasible region.
(b)Find the corner points of the feasible region.
(c) Find the maximum value of Z.

Answer

To solve the given Linear Programming Problem (LPP), we will follow the steps outlined in the question.

### Step 1: Identify the Objective Function and Constraints
We need to maximize the objective function:
\[ Z = 3x + 2y \]
subject to the constraints:
1. \( x + 2y \leq 10 \)
2. \( 3x + y \leq 15 \)
3. \( x \geq 0 \)
4. \( y \geq 0 \)

### Step 2: Convert Inequalities to Equations
To graph the constraints, we convert the inequalities into equations:
1. \( x + 2y = 10 \)
2. \( 3x + y = 15 \)

### Step 3: Find Intercepts for Each Constraint
**For the first constraint \( x + 2y = 10 \):**
- When \( x = 0 \):
\[ 2y = 10 \implies y = 5 \]
(Point: \( (0, 5) \))
- When \( y = 0 \):
\[ x = 10 \]
(Point: \( (10, 0) \))

**For the second constraint \( 3x + y = 15 \):**
- When \( x = 0 \):
\[ y = 15 \]
(Point: \( (0, 15) \))
- When \( y = 0 \):
\[ 3x = 15 \implies x = 5 \]
(Point: \( (5, 0) \))

### Step 4: Plot the Constraints on a Graph
1. Plot the points \( (0, 5) \) and \( (10, 0) \) for the first constraint.
2. Plot the points \( (0, 15) \) and \( (5, 0) \) for the second constraint.
3. Draw the lines for \( x + 2y = 10 \) and \( 3x + y = 15 \).
4. Shade the feasible region that satisfies all constraints, which lies in the first quadrant and below both lines.

### Step 5: Identify the Corner Points of the Feasible Region
The corner points of the feasible region can be found at the intersections of the lines and the axes:
1. \( (0, 0) \)
2. \( (0, 5) \)
3. \( (5, 0) \)
4. Intersection of \( x + 2y = 10 \) and \( 3x + y = 15 \).

**Finding the intersection:**
- From \( x + 2y = 10 \) (1)
- From \( 3x + y = 15 \) (2)

Multiply (1) by 3:
\[ 3x + 6y = 30 \]
Now subtract (2) from this:
\[ (3x + 6y) - (3x + y) = 30 - 15 \]
This simplifies to:
\[ 5y = 15 \implies y = 3 \]
Substituting \( y = 3 \) back into (1):
\[ x + 2(3) = 10 \implies x + 6 = 10 \implies x = 4 \]
So, the intersection point is \( (4, 3) \).

### Step 6: List the Corner Points
The corner points of the feasible region are:
1. \( (0, 0) \)
2. \( (0, 5) \)
3. \( (5, 0) \)
4. \( (4, 3) \)

### Step 7: Evaluate the Objective Function at Each Corner Point
Now we calculate the value of \( Z \) at each corner point:
1. At \( (0, 0) \):
\[ Z = 3(0) + 2(0) = 0 \]
2. At \( (0, 5) \):
\[ Z = 3(0) + 2(5) = 10 \]
3. At \( (5, 0) \):
\[ Z = 3(5) + 2(0) = 15 \]
4. At \( (4, 3) \):
\[ Z = 3(4) + 2(3) = 12 + 6 = 18 \]

### Step 8: Determine the Maximum Value of Z
The maximum value of \( Z \) occurs at the point \( (4, 3) \) where:
\[ Z = 18 \]

### Conclusion
The maximum value of \( Z \) is \( 18 \) at the corner point \( (4, 3) \).

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