The maximum value of Z =x +3y such that ` 2x +y le 20, x +2y le 20, x ge 0, y ge 0 ` is

Options

A. 10
B. 60
C. 30
D. None of these

Explanation

To solve the problem of maximizing \( Z = x + 3y \) subject to the constraints \( 2x + y \leq 20 \), \( x + 2y \leq 20 \), \( x \geq 0 \), and \( y \geq 0 \), we will follow these steps:

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

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

### Step 3: Find Intercepts
For each line, we find the x-intercept and y-intercept.

**For \( 2x + y = 20 \):**
- When \( x = 0 \): \( y = 20 \) (Point: \( (0, 20) \))
- When \( y = 0 \): \( 2x = 20 \) → \( x = 10 \) (Point: \( (10, 0) \))

**For \( x + 2y = 20 \):**
- When \( x = 0 \): \( 2y = 20 \) → \( y = 10 \) (Point: \( (0, 10) \))
- When \( y = 0 \): \( x = 20 \) (Point: \( (20, 0) \))

### Step 4: Plot the Lines
We plot the points \( (0, 20) \), \( (10, 0) \), \( (0, 10) \), and \( (20, 0) \) on the coordinate plane.

### Step 5: Determine Feasible Region
The feasible region is where all constraints overlap. We check the origin \( (0, 0) \) in both inequalities:
1. \( 2(0) + (0) \leq 20 \) → True
2. \( (0) + 2(0) \leq 20 \) → True

Thus, the feasible region is in the first quadrant and below the lines.

### Step 6: Find Intersection of the Lines
To find the intersection of the two lines:
1. From \( 2x + y = 20 \)
2. From \( x + 2y = 20 \)

We can solve these equations simultaneously. First, multiply the second equation by 2:
\[ 2x + 4y = 40 \]

Now we have:
1. \( 2x + y = 20 \)
2. \( 2x + 4y = 40 \)

Subtract the first equation from the second:
\[ 3y = 20 \]
\[ y = \frac{20}{3} \]

Substituting \( y \) back into the first equation:
\[ 2x + \frac{20}{3} = 20 \]
\[ 2x = 20 - \frac{20}{3} \]
\[ 2x = \frac{60}{3} - \frac{20}{3} = \frac{40}{3} \]
\[ x = \frac{20}{3} \]

Thus, the intersection point is \( \left(\frac{20}{3}, \frac{20}{3}\right) \).

### Step 7: Evaluate Objective Function at Corner Points
Now we evaluate \( Z = x + 3y \) at the corner points:
1. \( (0, 0) \): \( Z = 0 + 3(0) = 0 \)
2. \( (0, 10) \): \( Z = 0 + 3(10) = 30 \)
3. \( (10, 0) \): \( Z = 10 + 3(0) = 10 \)
4. \( \left(\frac{20}{3}, \frac{20}{3}\right) \):
\[ Z = \frac{20}{3} + 3\left(\frac{20}{3}\right) = \frac{20}{3} + \frac{60}{3} = \frac{80}{3} \approx 26.67 \]

### Step 8: Conclusion
The maximum value of \( Z \) occurs at the point \( (0, 10) \) and is \( Z = 30 \).

### Final Answer
The maximum value of \( Z \) is **30**.