Question

Q4: 3 marks A manufacturer has three machines I, Il and Il installed in his factory. Machines I and Il are capable of being operated for at most 12 hours whereas machine III must be operated for at least 5 hours a day. She produces only two items M and N each requiring the use of all the three machines. The number of hours required for producing 1 unit of each of M and N on the three machines are given in the following table: item Number of hours required on machines III П 3. She makes a profit of Rs 600 and Rs 400 on items M andN respectively. How many of each item should she produce so as to maximize her profit assuming that she can sell all the items that she produced? What will be the maximum profit? Hint: the objective function and the constraints for the decision variables for this optimization problem are as follows: Maximize the profit Z= 800x + 200 y Subject to the following constraints: x+2ys12 (constraint on Machine I) (1) 4x +3y 2 36 (constraint on Machine III).. (3) ....... x20, y 20 (non-negative constraint).. -(4) Find the feasible solution for the problem of a decorative item dealer whose LPP is to maximize profit function. (2 Marks) Determine the optimal solution to a LPP occurs at the corners of the feasible region. (1 Mark) 1. 2.

Answer 1

1. Let number of item M produced be x and that of item N be y.

Machine I (given):

Hours worked for product M = 1 hour

Hours worked for product N = 2 hours

Maximum hours available = 12

Thus constraint equation will be:

x +2y < 12

Machine II (given):

Hours worked for product M = 4 hours

Hours worked for product N = 3 hours

Thus constraint equation will be:

4r + 3y < 12

Maximum hours available = 12

Machine III (given):

Hours worked for product M = 0 hour

Hours worked for product N = 0 hours

Minimum hours = 5

Since both products don't use machine III, there will be no constraint.

Profit on item M = Rs 600

Profit on item N = Rs 400

We want to maximize the profit, hence the function used is maximize Z.

Thus profit maximization function will be:

Z = 600x + 400y

Combining all constraints:

Max Z = 600x + 400y

subject to:

x +2y < 12

4r + 3y < 12

x>0, y > 0

2. Please note: I have solved the equation by graphical method which is provided in the hint. The equation solved as per the given data in part 1 results in infeasible answers.

The graph is as below:

The lines intersect on points (6,3)

4 * 6/ اكلاتة 4 3 5 6 و 8 7 ،

Substituting x = 6 and y = 3 in the maximize function:

Max Z = 800x + 200y

Z = 800 * 6 + 200 * 3

Z = 4800 + 600

Z = 5400

Thus maximum profit that can be earned = Rs 5400