A producer has 20 and 10 unit of labour and capital respectively which he can use to produce two kinds of goods X and Y. To produce one unit of x, 2 units of capital and 1 unit of labour is required. To produce one unit of Y,3 of labour and 1 unit of capital is required. If X and Y are priced at Rs. 80 and Rs. 100 per unit respectively, how should the producer use his resources to maximise the total revenue?

Answer

To solve the problem of maximizing the total revenue for the producer using linear programming, we will follow these steps:

### Step 1: Define the Variables
Let:
- \( x \) = number of units of good X produced
- \( y \) = number of units of good Y produced

### Step 2: Formulate the Constraints
From the problem, we know the following about the resources required for production:
- To produce one unit of good X, 2 units of capital and 1 unit of labor are required.
- To produce one unit of good Y, 1 unit of capital and 3 units of labor are required.

Given the total resources available:
- Total labor available = 20 units
- Total capital available = 10 units

We can formulate the constraints based on these requirements:
1. **Labor Constraint**:
\[ 1x + 3y \leq 20 \]
2. **Capital Constraint**:
\[ 2x + 1y \leq 10 \]
3. **Non-negativity Constraints**:
\[ x \geq 0, \quad y \geq 0 \]

### Step 3: Formulate the Objective Function
The objective is to maximize the total revenue, which can be expressed as:
\[ Z = 80x + 100y \]
where \( 80 \) and \( 100 \) are the prices of goods X and Y respectively.

### Step 4: Graph the Constraints
To graph the constraints, we can convert the inequalities into equations to find the intercepts.

1. **For the labor constraint** \( x + 3y = 20 \):
- When \( x = 0 \), \( y = \frac{20}{3} \approx 6.67 \)
- When \( y = 0 \), \( x = 20 \)

2. **For the capital constraint** \( 2x + y = 10 \):
- When \( x = 0 \), \( y = 10 \)
- When \( y = 0 \), \( x = 5 \)

### Step 5: Identify the Feasible Region
Plot the lines on a graph and shade the feasible region that satisfies both constraints along with the non-negativity constraints. The feasible region will be bounded by the axes and the lines.

### Step 6: Find the Corner Points
The corner points of the feasible region can be found by solving the equations of the lines:
1. Intersection of \( x + 3y = 20 \) and \( 2x + y = 10 \):
- Solve the equations simultaneously to find the intersection point.

2. The other corner points will be the intercepts:
- \( (0, 0) \)
- \( (0, 10) \)
- \( (5, 0) \)
- Intersection point found above.

### Step 7: Evaluate the Objective Function at Each Corner Point
Calculate \( Z \) at each corner point:
1. \( Z(0, 0) = 80(0) + 100(0) = 0 \)
2. \( Z(0, 10) = 80(0) + 100(10) = 1000 \)
3. \( Z(5, 0) = 80(5) + 100(0) = 400 \)
4. \( Z \) at the intersection point (let's say it is \( (x_c, y_c) \)).

### Step 8: Determine the Maximum Revenue
Compare the values of \( Z \) calculated at the corner points. The maximum value will indicate the optimal production levels of goods X and Y.

### Final Answer
The producer should produce the quantities of goods X and Y that correspond to the corner point yielding the maximum revenue.