Q11 of 20 Page 12

If a young man drives his scooter at a speed of 25 km/hr, he has to spend Rs2 per km on petrol. If he drives the scooter at a speed of 40 km/hour, it produces air pollution and increases his expenditure on petrol to Rs 5 per km. He has a maximum of Rs100 to spend on petrol and travel a maximum distance in one hour time with less pollution. Express this problem as an LPP and solve it graphically. What value do you find here?

Let young man drives x km at a speed of 25 km/hr and y km at a speed of 40 km/hr. Clearly,


x, y 0


It is given that, he spends Rs 2 per km if he drives at a speed of 25 km/hr and Rs 5 per km if he drives at a speed of 40 km/hr. Therefore, money spent by him when he travelled x km and y km are Rs 2x and Rs 5y respectively.


It is given that he has a maximum of Rs 100 to spend.


Thus, 2x + 5y 100


Time spent by him when travelling with a speed of 25 km/hr = hr


Time spent by him when travelling with a speed of 40km/hr = hr


Also, the available time is 1 hour.



Or, 40x + 25y1000


The distance covered is Z = x + y which is to be maximized.


Thus, the mathematical formulation of the given linear programming problem is Max Z = x + y subject to


2x + 5y 100


40x + 25y1000


x, y 0


First we will convert inequations as follows:


2x + 5y = 100


40x + 25y = 1000


x = 0 and y = 0.


The region represented by 2x + 5y 100


The line 2x + 5y = 100 meets the coordinate axes at A(50,0) and B(0,20) respectively. By joining these points, we obtain the line 2x + 5y = 100. Clearly (0, 0) satisfies the 2x + 5y = 100. So, the region which contains the origin represents the solution set of the inequation 2x + 5y 100


The region represented by 40x + 25y 1000


The line 40x + 25y = 1000 meets the coordinate axes at C(25,0) and D(0,40) respectively. By joining these points, we obtain the line 2x + y = 12. Clearly (0, 0) satisfies the 40x + 25y = 1000. So, the region which contains the origin represents the solution set of the inequation 40x + 25y 1000


The region represented by x 0, y 0 :


Since every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by the inequations x 0 and y 0.


The feasible region determined by the system of constraints


2x + 5y 100, 40x + 25y1000, x 0 and y 0 are as follows


1.jpg


The corner points are O(0,0), B(0,20), E, and C(25,0). The value of Z at these corner points are as follows:



The maximum value of Z is 30 which is attained at E.


Thus, the maximum distance travelled by the young man is 30 kms, if he drives km at a speed of 25 km/hr and km at a speed of 40 km/hr.

More from this chapter

All 20 →
9

A firm manufactures 3 products A, B and C. The profits are Rs. 3, Rs. 2 and Rs. 4 respectively. The firm has 2 machines and below is the required processing time in minutes for each machine on each product.


Machines M1 and M2 have 2000 machine minutes respectively. The firm must manufacture 100 A’s, 200 B’s and 50 C’s but not more than 150 A’s. Set up a LPP to maximize the profit.

 10

Solve each of the following linear programming problems by graphical method.

Maximize Z = 50x + 30y


Subject to :


2x + y ≤ 18


3x + 2y ≤ 34


x, y ≥ 0

 12

A factory manufactures two types of screws, A and B, each type requiring the use of two machines - an automatic and a hand - operated. It takes 4 minute on the automatic and 6 minutes on the hand - operated machines to manufacture a package of screws ‘A’, while it takes 6 minutes on the automatic and 3 minutes on the hand - operated machine to manufacture a package of screws ‘B’. Each machine is available for at most 4 hours on any day. The manufacturer can sell a package of screws ‘A’ at a profit of 70 P and screws ‘B’ at a profit of ₹ 1. Assuming that he can sell all the screws he can manufacture, how many packages of each type should the factory owner produce in a day in order to maximize his profit? Determine the maximum profit.

 13

A manufacturer makes two types A and B of tea - cups. Three machines are needed for the manufacture and the time in minutes required for each cup on the machines is given below :


Each machine is available for a maximum of 6 hours per day. If the profit on each cup A is 75 paise and that on each cup B is 50 paise, show that 15 tea - cups of type A and 30 of type B should be manufactured in a day to get the maximum profit.