A toy company manufacutes two types of dolls, A and B. Each doll of type B takes twice as long to produce as one of type. A and the company would have time to make a maxium of 2000 prr day, if it produces only type.A the supply of plastic is sufficient to produce 1500 dolls per day (both A and B combined). Type B requires a fancy dress of which there are only 600 per day available. If the company makes profits of Rs.3 and Rs5 per doll respectively on dolls A and B, how many of each should be produced per day in order to mazimize the profit? Also, find the maximum profit.

Explanation

To solve the problem of maximizing the profit for the toy company manufacturing dolls A and B, we will follow these steps:

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

### Step 2: Formulate the Objective Function
The profit from dolls A and B is given as:
- Profit from A = \( 3x \) (since each doll A gives a profit of Rs. 3)
- Profit from B = \( 5y \) (since each doll B gives a profit of Rs. 5)

Thus, the total profit \( Z \) to be maximized is:
\[ Z = 3x + 5y \]

### Step 3: Set Up the Constraints
From the problem statement, we have the following constraints:

1. Time constraint: Each doll of type B takes twice as long to produce as one of type A. Therefore, if only type A is produced, the maximum production is 2000 dolls per day. This gives us:
\[ x + 2y \leq 2000 \]

2. Plastic supply constraint: The total production of both types of dolls cannot exceed 1500 dolls per day:
\[ x + y \leq 1500 \]

3. Fancy dress constraint: The production of type B dolls is limited to 600 dolls per day:
\[ y \leq 600 \]

4. Non-negativity constraints: The number of dolls produced cannot be negative:
\[ x \geq 0, \quad y \geq 0 \]

### Step 4: Graph the Constraints
To find the feasible region, we need to graph the constraints:

1. For \( x + 2y = 2000 \):
- If \( x = 0 \), \( y = 1000 \) (point A)
- If \( y = 0 \), \( x = 2000 \) (point B)

2. For \( x + y = 1500 \):
- If \( x = 0 \), \( y = 1500 \) (point C)
- If \( y = 0 \), \( x = 1500 \) (point D)

3. For \( y = 600 \):
- This is a horizontal line at \( y = 600 \).

### Step 5: Find the Intersection Points
To find the vertices of the feasible region, we need to find the intersection points of the lines:

1. **Intersection of \( x + 2y = 2000 \) and \( x + y = 1500 \)**:
- Solve the equations:
\[ x + 2y = 2000 \quad (1) \]
\[ x + y = 1500 \quad (2) \]
- Subtract (2) from (1):
\[ (x + 2y) - (x + y) = 2000 - 1500 \implies y = 500 \]
- Substitute \( y = 500 \) into (2):
\[ x + 500 = 1500 \implies x = 1000 \]
- Intersection point: \( (1000, 500) \)

2. **Intersection of \( x + y = 1500 \) and \( y = 600 \)**:
- Substitute \( y = 600 \) into \( x + y = 1500 \):
\[ x + 600 = 1500 \implies x = 900 \]
- Intersection point: \( (900, 600) \)

3. **Intersection of \( x + 2y = 2000 \) and \( y = 600 \)**:
- Substitute \( y = 600 \) into \( x + 2y = 2000 \):
\[ x + 1200 = 2000 \implies x = 800 \]
- Intersection point: \( (800, 600) \)

### Step 6: Identify the Feasible Region
The feasible region is bounded by the points:
- \( (0, 0) \)
- \( (1000, 500) \)
- \( (900, 600) \)
- \( (800, 600) \)
- \( (1500, 0) \)

### Step 7: Evaluate the Objective Function at Each Vertex
Now we will evaluate \( Z = 3x + 5y \) at each vertex:

1. At \( (0, 0) \):
\[ Z = 3(0) + 5(0) = 0 \]

2. At \( (1000, 500) \):
\[ Z = 3(1000) + 5(500) = 3000 + 2500 = 5500 \]

3. At \( (900, 600) \):
\[ Z = 3(900) + 5(600) = 2700 + 3000 = 5700 \]

4. At \( (800, 600) \):
\[ Z = 3(800) + 5(600) = 2400 + 3000 = 5400 \]

### Step 8: Determine the Maximum Profit
The maximum profit occurs at the point \( (900, 600) \) with:
\[ \text{Maximum Profit} = 5700 \]

### Conclusion
The company should produce:
- \( 900 \) dolls of type A
- \( 600 \) dolls of type B

**Maximum Profit = Rs. 5700**