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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?

1 answer

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.