Question

A manufacturer of furniture makes two products .chairs and tables. processing of this product is done on two machines A and B .A chair requires 2 hours on machine A and 6 hours on machine B.A table requires 5 hours on machine A and no time on machine B.there are 16 hours of time per day available on machine A and 30 hours on machine B.profit gained by the manufacturer from a chair and a table is br 2 and br 10 respectively. what should be the daily production of each of two products? and find the optimal value using graphical method?

Answer 1

To solve this problem, we can use linear programming. Let's define the variables first:

- Let \( x \) be the number of chairs produced.
- Let \( y \) be the number of tables produced.

### Given:
1. **Time constraints:**
- Each chair requires 2 hours on machine A and 6 hours on machine B.
- Each table requires 5 hours on machine A and no time on machine B.
- There are 16 hours available on machine A and 30 hours available on machine B.

2. **Profit:**
- Profit from each chair = 2 BR
- Profit from each table = 10 BR

### Constraints:
Based on the processing times and availability, we can create the following inequalities for the constraints:

1. For machine A (time constraint):
\[
2x + 5y \leq 16
\]

2. For machine B (time constraint):
\[
6x + 0y \leq 30 \quad \Rightarrow \quad 6x \leq 30 \quad \Rightarrow \quad x \leq 5
\]

3. Non-negativity constraints:
\[
x \geq 0
\]
\[
y \geq 0
\]

### Objective Function:
We want to maximize the profit \( P \):
\[
P = 2x + 10y
\]

### Graphical Method:
We will graph the inequalities to find the feasible region.

1. **Graph \( 2x + 5y \leq 16 \):**
- To find the intercepts, set \( x = 0 \):
\[
5y = 16 \quad \Rightarrow \quad y = 3.2 \quad \text{(Point: }(0, 3.2)\text{)}
\]
- Set \( y = 0 \):
\[
2x = 16 \quad \Rightarrow \quad x = 8 \quad \text{(Point: }(8, 0)\text{)}
\]

2. **Graph \( x \leq 5 \):**
- This is a vertical line at \( x = 5 \).

3. **Non-negativity constraints:**
- All \( x \geq 0 \) and \( y \geq 0 \).

### Feasible Region:
The feasible region is the area where all constraints overlap, bounded by the intercepts calculated and the axes.

### Corners of the Feasible Region:
1. \( (0, 0) \)
2. \( (0, 3.2) \)
3. \( (5, 0) \)
4. \( (5, ...)\) from \( 2(5) + 5y = 16 \):
\[
10 + 5y = 16 \quad \Rightarrow \quad 5y = 6 \quad \Rightarrow \quad y = 1.2 \quad \text{(Point: }(5, 1.2)\text{)}
\]

### Evaluate the Objective Function at Each Corner:
- At \( (0, 0) \): \( P = 2(0) + 10(0) = 0 \)
- At \( (0, 3.2) \): \( P = 2(0) + 10(3.2) = 32 \)
- At \( (5, 0) \): \( P = 2(5) + 10(0) = 10 \)
- At \( (5, 1.2) \): \( P = 2(5) + 10(1.2) = 10 + 12 = 22 \)

### Optimal Solution:
The maximum profit occurs at \( (0, 3.2) \) with a profit of 32 BR. Thus, the recommended daily production should be:
- **Chairs:** 0
- **Tables:** 3.2 (rounded down to 3, if needed for integer solutions)

So, the optimal production plan is to produce 3 tables and no chairs for maximum profit given the conditions.