Advertisements
Advertisements
Question
A man rides his motorcycle at the speed of 50 km/hour. He has to spend Rs 2 per km on petrol. If he rides it at a faster speed of 80 km/hour, the petrol cost increases to Rs 3 per km. He has atmost Rs 120 to spend on petrol and one hour’s time. He wishes to find the maximum distance that he can travel. Express this problem as a linear programming problem
Advertisements
Solution
Let the man covers x km on his motorcycle at the speed of 50 km/hr and covers y km at the speed of 80 km/hr.
So, cost of petrol = 2x + 3y
The man has to spend ₹ 120 atmost on petrol
∴ 2x + 3y ≤ 120 ......(i)
Now, the man has only 1 hr time
∴ `x/50 + y/80 ≤ 1`
⇒ 8x + 5y ≤ 400 .....(ii)
x ≥ 0, y ≥ 0
To have maximum distance Z = x + y.
Hence, the required LPP to travel maximum distance is maximise Z = x + y
Subject to the constraints 2x + 3y ≤ 120, 8x + 5y ≤ 400, x ≥ 0, y ≥ 0.
APPEARS IN
RELATED QUESTIONS
Two tailors, A and B, earn Rs 300 and Rs 400 per day respectively. A can stitch 6 shirts and 4 pairs of trousers while B can stitch 10 shirts and 4 pairs of trousers per day. To find how many days should each of them work and if it is desired to produce at least 60 shirts and 32 pairs of trousers at a minimum labour cost, formulate this as an LPP
Solve the following Linear Programming Problems graphically:
Maximise Z = 3x + 4y
subject to the constraints : x + y ≤ 4, x ≥ 0, y ≥ 0.
Solve the following Linear Programming Problems graphically:
Maximise Z = 3x + 2y
subject to x + 2y ≤ 10, 3x + y ≤ 15, x, y ≥ 0.
Solve the following Linear Programming Problems graphically:
Minimise Z = x + 2y
subject to 2x + y ≥ 3, x + 2y ≥ 6, x, y ≥ 0.
Show that the minimum of Z occurs at more than two points.
Maximise Z = x + y, subject to x – y ≤ –1, –x + y ≤ 0, x, y ≥ 0.
A manufacturer makes two types of toys A and B. Three machines are needed for this purpose and the time (in minutes) required for each toy on the machines is given below:
| Type of toy | Machines | ||
| I | II | III | |
| A | 12 | 18 | 6 |
| B | 6 | 0 | 9 |
Each machine is available for a maximum of 6 hours per day. If the profit on each toy of type A is Rs 7.50 and that on each toy of type B is Rs 5, show that 15 toys of type A and 30 of type B should be manufactured in a day to get maximum profit.
An aeroplane can carry a maximum of 200 passengers. A profit of Rs 1000 is made on each executive class ticket and a profit of Rs 600 is made on each economy class ticket. The airline reserves at least 20 seats for executive class. However, at least 4 times as many passengers prefer to travel by economy class than by the executive class. Determine how many tickets of each type must be sold in order to maximize the profit for the airline. What is the maximum profit?
Determine the maximum value of Z = 11x + 7y subject to the constraints : 2x + y ≤ 6, x ≤ 2, x ≥ 0, y ≥ 0.
Maximise Z = 3x + 4y, subject to the constraints: x + y ≤ 1, x ≥ 0, y ≥ 0
Maximise the function Z = 11x + 7y, subject to the constraints: x ≤ 3, y ≤ 2, x ≥ 0, y ≥ 0.
Determine the maximum value of Z = 3x + 4y if the feasible region (shaded) for a LPP is shown in Figure
The feasible region for a LPP is shown in Figure. Find the minimum value of Z = 11x + 7y
Refer to Exercise 7 above. Find the maximum value of Z.
Refer to question 13. Solve the linear programming problem and determine the maximum profit to the manufacturer
Refer to question 15. Determine the maximum distance that the man can travel.
A manufacturer produces two Models of bikes-Model X and Model Y. Model X takes a 6 man-hours to make per unit, while Model Y takes 10 man-hours per unit. There is a total of 450 man-hour available per week. Handling and Marketing costs are Rs 2000 and Rs 1000 per unit for Models X and Y respectively. The total funds available for these purposes are Rs 80,000 per week. Profits per unit for Models X and Y are Rs 1000 and Rs 500, respectively. How many bikes of each model should the manufacturer produce so as to yield a maximum profit? Find the maximum profit.
The corner points of the feasible region determined by the system of linear constraints are (0, 0), (0, 40), (20, 40), (60, 20), (60, 0). The objective function is Z = 4x + 3y ______.
Compare the quantity in Column A and Column B
| Column A | Column B |
| Maximum of Z | 325 |
The feasible solution for a LPP is shown in Figure. Let Z = 3x – 4y be the objective function. Minimum of Z occurs at ______.
Refer to Question 30. Minimum value of F is ______.
If the feasible region for a LPP is ______ then the optimal value of the objective function Z = ax + by may or may not exist.
In a LPP if the objective function Z = ax + by has the same maximum value on two corner points of the feasible region, then every point on the line segment joining these two points give the same ______ value.
A corner point of a feasible region is a point in the region which is the ______ of two boundary lines.
If the feasible region for a LPP is unbounded, maximum or minimum of the objective function Z = ax + by may or may not exist.
Maximum value of the objective function Z = ax + by in a LPP always occurs at only one corner point of the feasible region.
Based on the given shaded region as the feasible region in the graph, at which point(s) is the objective function Z = 3x + 9y maximum?
Z = 7x + y, subject to 5x + y ≥ 5, x + y ≥ 3, x ≥ 0, y ≥ 0. The minimum value of Z occurs at ____________.
In linear programming, optimal solution ____________.
A maximum or a minimum may not exist for a linear programming problem if ____________.
In Corner point method for solving a linear programming problem, one finds the feasible region of the linear programming problem, determines its corner points, and evaluates the objective function Z = ax + by at each corner point. Let M and m respectively be the largest and smallest values at corner points. In case feasible region is unbounded, M is the maximum value of the objective function if ____________.
Maximize Z = 6x + 4y, subject to x ≤ 2, x + y ≤ 3, -2x + y ≤ 1, x ≥ 0, y ≥ 0.
Maximize Z = 10 x1 + 25 x2, subject to 0 ≤ x1 ≤ 3, 0 ≤ x2 ≤ 3, x1 + x2 ≤ 5.
Z = 6x + 21 y, subject to x + 2y ≥ 3, x + 4y ≥ 4, 3x + y ≥ 3, x ≥ 0, y ≥ 0. The minimum value of Z occurs at ____________.
Maximize Z = 10×1 + 25×2, subject to 0 ≤ x1 ≤ 3, 0 ≤ x2 ≤ 3, x1 + x2 ≤ 5.
