In the manufacturer of electronic circuits has a stock of 200 resistors, 120 transistors and 150 capacitors and is required to produce two types of circuits A and B. Type A requires 20 resistors, 10 transistors and 10 capacitors. Type B requires 10 resistros, 20 transistors and 30 capacitors, If the profit on type A circuit is 50 and that on type B circuit is 60 formulate this problem as LPP, so that hte manufacturer can maximise his profit.

Explanation

To formulate the given problem as a Linear Programming Problem (LPP), we will follow these steps:

### Step 1: Define the Variables
Let:
- \( x \) = number of units of type A circuits produced
- \( y \) = number of units of type B circuits produced

### Step 2: Define the Objective Function
The objective is to maximize the profit. The profit from type A circuits is \( 50x \) and from type B circuits is \( 60y \). Therefore, the objective function can be written as:
\[ Z = 50x + 60y \]

### Step 3: Define the Constraints
We need to consider the availability of resistors, transistors, and capacitors.

1. **Resistors Constraint**:
- Type A requires 20 resistors per unit, and type B requires 10 resistors per unit.
- Total available resistors = 200.
\[ 20x + 10y \leq 200 \]

2. **Transistors Constraint**:
- Type A requires 10 transistors per unit, and type B requires 20 transistors per unit.
- Total available transistors = 120.
\[ 10x + 20y \leq 120 \]

3. **Capacitors Constraint**:
- Type A requires 10 capacitors per unit, and type B requires 30 capacitors per unit.
- Total available capacitors = 150.
\[ 10x + 30y \leq 150 \]

### Step 4: Non-negativity Constraints
Since the number of circuits produced cannot be negative, we have:
\[ x \geq 0, \quad y \geq 0 \]

### Final Formulation
The Linear Programming Problem can be summarized as follows:

**Maximize**:
\[ Z = 50x + 60y \]

**Subject to**:
\[ \begin{align*} 20x + 10y & \leq 200 \quad \text{(Resistors)} \\ 10x + 20y & \leq 120 \quad \text{(Transistors)} \\ 10x + 30y & \leq 150 \quad \text{(Capacitors)} \\ x & \geq 0 \\ y & \geq 0 \end{align*} \]