Optimum solution of the L.P.P. :
Maximize `z= 3x+5y`
subject to `3x+2y le 18, x le 4, y le 6, x ge 0, y ge 0` is

Options

A. `x=2, y =0, z=6`
B. `x=2, y=6, z=36`
C. `x=4, y=3, z=27`
D. `x=4, y=6, z=42`

Answer

To solve the given Linear Programming Problem (LPP) step-by-step, we will follow these steps:

### Step 1: Define the Objective Function
We need to maximize the objective function:
\[ z = 3x + 5y \]

### Step 2: Identify the Constraints
The constraints given are:
1. \( 3x + 2y \leq 18 \)
2. \( x \leq 4 \)
3. \( y \leq 6 \)
4. \( x \geq 0 \)
5. \( y \geq 0 \)

### Step 3: Convert Inequalities to Equations
To graph the constraints, we convert the inequalities into equations:
1. \( 3x + 2y = 18 \)
2. \( x = 4 \)
3. \( y = 6 \)

### Step 4: Find Intercepts for the First Constraint
For the equation \( 3x + 2y = 18 \):
- To find the x-intercept, set \( y = 0 \):
\[ 3x = 18 \implies x = 6 \]
- To find the y-intercept, set \( x = 0 \):
\[ 2y = 18 \implies y = 9 \]

### Step 5: Plot the Constraints
On a graph:
- Plot the line \( 3x + 2y = 18 \) with intercepts (6,0) and (0,9).
- Draw the vertical line \( x = 4 \).
- Draw the horizontal line \( y = 6 \).
- Also, consider the axes where \( x \geq 0 \) and \( y \geq 0 \).

### Step 6: Identify the Feasible Region
The feasible region is where all the constraints overlap. This area will be bounded by the lines we plotted and will lie in the first quadrant (since \( x \geq 0 \) and \( y \geq 0 \)).

### Step 7: Determine the Corner Points
The corner points of the feasible region can be found by solving the equations of the lines:
1. Intersection of \( 3x + 2y = 18 \) and \( y = 6 \):
\[ 3x + 2(6) = 18 \implies 3x + 12 = 18 \implies 3x = 6 \implies x = 2 \implies (2, 6) \]
2. Intersection of \( 3x + 2y = 18 \) and \( x = 4 \):
\[ 3(4) + 2y = 18 \implies 12 + 2y = 18 \implies 2y = 6 \implies y = 3 \implies (4, 3) \]
3. The other corner points are:
- \( (0, 0) \)
- \( (0, 6) \)
- \( (4, 0) \)

### Step 8: Evaluate the Objective Function at Each Corner Point
Now we evaluate \( z = 3x + 5y \) at each corner point:
1. At \( (0, 0) \): \( z = 3(0) + 5(0) = 0 \)
2. At \( (0, 6) \): \( z = 3(0) + 5(6) = 30 \)
3. At \( (4, 0) \): \( z = 3(4) + 5(0) = 12 \)
4. At \( (2, 6) \): \( z = 3(2) + 5(6) = 6 + 30 = 36 \)
5. At \( (4, 3) \): \( z = 3(4) + 5(3) = 12 + 15 = 27 \)

### Step 9: Determine the Maximum Value
The maximum value of \( z \) occurs at the point \( (2, 6) \):
\[ z_{\text{max}} = 36 \]

### Conclusion
The optimum solution of the LPP is:
- \( x = 2 \)
- \( y = 6 \)
- Maximum value of \( z = 36 \)

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