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5.3 Q-8

Question Statement

A dealer wishes to purchase fans and sewing machines. He has Rs. 5760 to invest and has space for a maximum of 20 items. A fan costs Rs. 360 and a sewing machine costs Rs. 240. The dealer expects to make a profit of Rs. 22 per fan and Rs. 18 per sewing machine. The dealer aims to maximize his profit by determining how many fans and sewing machines he should buy. The objective is to find the optimal number of each item to purchase, based on the constraints.


Background and Explanation

To solve this problem, we are dealing with a Linear Programming Problem (LPP) where we want to maximize a profit function under certain constraints. The dealer’s profit from buying fans and sewing machines is expressed as:

f(x,y)=22x+18yf(x, y) = 22x + 18y

Where:

  • xx is the number of fans purchased.
  • yy is the number of sewing machines purchased.

The problem is subject to the following constraints:

  1. Total cost should not exceed Rs. 5760.
  2. Total number of items must be at most 20.
  3. Both xx and yy must be non-negative.

Solution

We begin by expressing the constraints in terms of equations:

  1. The cost of the fans and sewing machines should not exceed the total investment of Rs. 5760: 360x+240y5760360x + 240y \leq 5760

  2. The total number of items purchased should not exceed 20: x+y20x + y \leq 20

Also, we know:

  • x0x \geq 0
  • y0y \geq 0

We can rewrite the first constraint as: 120(3x+2y)5760120(3x + 2y) \leq 5760

This simplifies to: 3x+2y483x + 2y \leq 48

So, the system of inequalities becomes:

  • 3x+2y483x + 2y \leq 48
  • x+y20x + y \leq 20
  • x0x \geq 0
  • y0y \geq 0

Now, we will plot the equations and find the feasible region.

Step 1: Find the Points of Intersection

For 3x+2y=483x + 2y = 48:

  • Setting x=0x = 0 gives y=24y = 24, so the point is (0,24)(0, 24).
  • Setting y=0y = 0 gives x=16x = 16, so the point is (16,0)(16, 0).

For x+y=20x + y = 20:

  • Setting x=0x = 0 gives y=20y = 20, so the point is (0,20)(0, 20).
  • Setting y=0y = 0 gives x=20x = 20, so the point is (20,0)(20, 0).

Checking the Feasibility of Points:

  • For x=0,y=0x = 0, y = 0, both constraints are satisfied.
  • Points like (0,20)(0, 20), (16,0)(16, 0), and (8,12)(8, 12) are feasible as they satisfy all constraints.

Step 2: Graphical Representation

The graph of these equations shows the feasible region, which is the shaded area in the first quadrant. We will now find the corner points to check the profit at each of them.


Key Formulas or Methods Used

  • Objective Function: f(x,y)=22x+18yf(x, y) = 22x + 18y

  • Constraints:

    • 360x+240y5760360x + 240y \leq 5760
    • x+y20x + y \leq 20
    • 3x+2y483x + 2y \leq 48
    • x0,y0x \geq 0, y \geq 0

Summary of Steps

  1. Define the objective function to maximize: f(x,y)=22x+18yf(x, y) = 22x + 18y.
  2. Write the constraints based on the cost and space limits.
  3. Solve the system of inequalities to find the points of intersection.
  4. Check the corner points: (8,12)(8, 12), (0,20)(0, 20), (0,0)(0, 0), and (16,0)(16, 0).
  5. Evaluate the profit function at each corner point to find the maximum profit.
  6. The optimal solution is to purchase 8 fans and 12 sewing machines, resulting in a maximum profit of Rs. 392.