Solve the linear programming problem. Maximise Z = x + 2y
Subject to constraints `x - y le 10, 2x +3y le 20 ` and ` x ge 0, y ge 0 `

Options

A. Z=10
B. Z=30
C. Z=40
D. None of these

Explanation

To solve the linear programming problem, we need to maximize the objective function \( Z = x + 2y \) subject to the given constraints. Let's go through the steps systematically.

### Step 1: Identify the Constraints
The constraints given are:
1. \( x - y \leq 10 \)
2. \( 2x + 3y \leq 20 \)
3. \( x \geq 0 \)
4. \( y \geq 0 \)

### Step 2: Convert Inequalities to Equations
To find the boundary lines, we convert the inequalities into equations:
1. \( x - y = 10 \)
2. \( 2x + 3y = 20 \)

### Step 3: Find the Intercepts
For the first equation \( x - y = 10 \):
- When \( x = 0 \): \( 0 - y = 10 \) → \( y = -10 \) (not valid since \( y \geq 0 \))
- When \( y = 0 \): \( x - 0 = 10 \) → \( x = 10 \) (point is \( (10, 0) \))

For the second equation \( 2x + 3y = 20 \):
- When \( x = 0 \): \( 2(0) + 3y = 20 \) → \( y = \frac{20}{3} \) (point is \( (0, \frac{20}{3}) \))
- When \( y = 0 \): \( 2x + 3(0) = 20 \) → \( x = 10 \) (point is \( (10, 0) \))

### Step 4: Find the Intersection of the Lines
To find the intersection of the lines \( x - y = 10 \) and \( 2x + 3y = 20 \), we can solve these equations simultaneously.

From \( x - y = 10 \), we can express \( y \) in terms of \( x \):
\[ y = x - 10 \]

Substituting \( y \) in the second equation:
\[ 2x + 3(x - 10) = 20 \]
\[ 2x + 3x - 30 = 20 \]
\[ 5x - 30 = 20 \]
\[ 5x = 50 \]
\[ x = 10 \]

Substituting \( x = 10 \) back to find \( y \):
\[ y = 10 - 10 = 0 \]
So, the intersection point is \( (10, 0) \).

### Step 5: Identify the Feasible Region
The feasible region is determined by the constraints:
- The line \( x - y = 10 \) is above the line.
- The line \( 2x + 3y = 20 \) is below the line.
- The region is also restricted to the first quadrant due to \( x \geq 0 \) and \( y \geq 0 \).

The vertices of the feasible region are:
1. \( (0, 0) \)
2. \( (10, 0) \)
3. \( (0, \frac{20}{3}) \)

### Step 6: Evaluate the Objective Function at Each Vertex
Now, we evaluate \( Z = x + 2y \) at each vertex:
1. At \( (0, 0) \): \( Z = 0 + 2(0) = 0 \)
2. At \( (10, 0) \): \( Z = 10 + 2(0) = 10 \)
3. At \( (0, \frac{20}{3}) \): \( Z = 0 + 2(\frac{20}{3}) = \frac{40}{3} \)

### Step 7: Determine the Maximum Value
Comparing the values:
- \( Z(0, 0) = 0 \)
- \( Z(10, 0) = 10 \)
- \( Z(0, \frac{20}{3}) = \frac{40}{3} \approx 13.33 \)

The maximum value of \( Z \) occurs at the point \( (0, \frac{20}{3}) \) with \( Z = \frac{40}{3} \).

### Final Answer
The maximum value of \( Z \) is \( \frac{40}{3} \) at the point \( (0, \frac{20}{3}) \).

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