Let’s take x an y to be the number of models of bike produced by the manufacturer.
From the question we have,
Model x takes 6 man-hours to make per unit
Model y takes 10 man-hours to make per unit
Total man-hours available = 450
So, 6x + 10y ≤ 450 ⇒ 3x + 5y ≤ 225 ….. (i)
The handling and marketing cost of model x and y are Rs 2000 and Rs 1000 respectively.
And, the total funds available is Rs 80,000 per week
So, 2000x + 1000y ≤ 80000 ⇒ 2x + y ≤ 80 … (ii)
And, x ≥ 0, y ≥ 0
Now, the total profit (Z) per unit of models x and y are Rs 1000 and Rs 500 repectively
⇒ Z = 1000x + 500y
Hence, the required LPP is
Maximize Z = 1000x + 500y subject to the constraints
3x + 5y ≤ 225, 2x + y ≤ 80 and x ≥ 0, y ≥ 0
Now, let’s construct a constrain table for the above:
Table for (i)
Next, on solving equation (i) and (ii) we get
x = 25 and y = 30
After plotting all the constraint equations, we observe that the feasible region is OABC, whose corner points are O(0, 0), A(40, 0), B(25, 30) and C(0, 45).
On evaluating the value of Z, we get
Therefore, from the above table it’s seen that the maximum profit is Rs 40,000.
The maximum profit can be achieved by producing 25 bikes of model x and 30 bikes of model Y or by producing 40 bikes of model x.