Advertisements
Advertisements
Question
To maintain his health a person must fulfil certain minimum daily requirements for several kinds of nutrients. Assuming that there are only three kinds of nutrients-calcium, protein and calories and the person's diet consists of only two food items, I and II, whose price and nutrient contents are shown in the table below:
| Food I (per lb) |
Food II (per lb) |
Minimum daily requirement for the nutrient |
||||
| Calcium | 10 | 5 | 20 | |||
| Protein | 5 | 4 | 20 | |||
| Calories | 2 | 6 | 13 | |||
| Price (Rs) | 60 | 100 |
What combination of two food items will satisfy the daily requirement and entail the least cost? Formulate this as a LPP.
Advertisements
Solution
Let the person takes x lbs and y lbs of food I and II respectively that were taken in the diet.
Since, per lb of food I costs Rs 60 and that of food II costs Rs 100.
Therefore, x lbs of food I costs Rs 60x and y lbs of food II costs Rs 100y.
Total cost per day = Rs (60x + 100y)
Let Z denote the total cost per day
Then, Z = 60x + 100y
Total amount of calcium in the diet is \[10x + 5y\]
Since, each lb of food I contains 10 units of calcium.Therefore, x lbs of food I contains 10x units of calcium.
Each lb of food II contains 5 units of calciu.So,y lbs of food II contains 5y units of calcium.
Thus, x lbs of food I and y lbs of food II contains 10x + 5y units of calcium.
But, the minimum requirement is 20 lbs of calcium.
Each lb of food II contains 4 units of protein.So,y lbs of food II contains 4y units of protein.
Thus, x lbs of food I and y lbs of food II contains 5x + 4y units of protein.
But, the minimum requirement is 20 lbs of protein.
Each lb of food II contains units of calories.So,y lbs of food II contains 6y units of calories.
Thus, x lbs of food I and y lbs of food II contains
But, the minimum requirement is 13 lbs of calories.
So,
Min Z = 60x + 100y
subject to
\[5x + 4y \geq 20\]
\[2x + 6y \geq 13\]
\[x, y \geq 0\]
APPEARS IN
RELATED QUESTIONS
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.
Show that the minimum of Z occurs at more than two points.
Minimise and Maximise Z = 5x + 10 y
subject to x + 2y ≤ 120, x + y ≥ 60, x – 2y ≥ 0, x, y ≥ 0.
Show that the minimum of Z occurs at more than two points.
Minimise and Maximise Z = x + 2y
subject to x + 2y ≥ 100, 2x – y ≤ 0, 2x + y ≤ 200; x, y ≥ 0.
Refer to Example 9. How many packets of each food should be used to maximize the amount of vitamin A in the diet? What is the maximum amount of vitamin A in the diet?
A farmer mixes two brands P and Q of cattle feed. Brand P, costing Rs 250 per bag contains 3 units of nutritional element A, 2.5 units of element B and 2 units of element C. Brand Q costing Rs 200 per bag contains 1.5 units of nutritional elements A, 11.25 units of element B, and 3 units of element C. The minimum requirements of nutrients A, B and C are 18 units, 45 units and 24 units respectively. Determine the number of bags of each brand which should be mixed in order to produce a mixture having a minimum cost per bag? What is the minimum cost of the mixture per bag?
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?
A small firm manufactures necklaces and bracelets. The total number of necklaces and bracelets that it can handle per day is at most 24. It takes one hour to make a bracelet and half an hour to make a necklace. The maximum number of hours available per day is 16. If the profit on a necklace is Rs 100 and that on a bracelet is Rs 300. Formulate on L.P.P. for finding how many of each should be produced daily to maximize the profit?
It is being given that at least one of each must be produced.
The minimum value of the objective function Z = ax + by in a linear programming problem always occurs at only one corner point of the feasible region
Determine the maximum value of Z = 11x + 7y subject to the constraints : 2x + y ≤ 6, x ≤ 2, x ≥ 0, y ≥ 0.
Feasible region (shaded) for a LPP is shown in Figure. Maximise Z = 5x + 7y.
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.
Refer to question 15. Determine the maximum distance that the man can travel.
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 |
Refer to Question 27. (Maximum value of Z + Minimum value of Z) is equal to ______.
Refer to Question 30. Minimum value of F is ______.
Refer to Question 32, Maximum of F – Minimum of F = ______.
In a LPP, the linear inequalities or restrictions on the variables are called ____________.
In a LPP, the objective function is always ______.
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 feasible region of a system of linear inequalities is said to be ______ if it can be enclosed within a circle.
The feasible region for an LPP is always a ______ polygon.
If the feasible region for a LPP is unbounded, maximum or minimum of the objective function Z = ax + by may or may not exist.
In a linear programming problem, the constraints on the decision variables x and y are x − 3y ≥ 0, y ≥ 0, 0 ≤ x ≤ 3. The feasible region:
Objective function of a linear programming problem is ____________.
The maximum value of the object function Z = 5x + 10 y subject to the constraints x + 2y ≤ 120, x + y ≥ 60, x - 2y ≥ 0, x ≥ 0, y ≥ 0 is ____________.
Z = 7x + y, subject to 5x + y ≥ 5, x + y ≥ 3, x ≥ 0, y ≥ 0. The minimum value of Z occurs at ____________.
A linear programming problem is one that is concerned with ____________.
In linear programming infeasible solutions
In linear programming, optimal solution ____________.
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. If M and m respectively be the largest and smallest values at corner points then ____________.
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 ____________.
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 the feasible region is unbounded, m is the minimum value of the objective function.
Maximize Z = 7x + 11y, subject to 3x + 5y ≤ 26, 5x + 3y ≤ 30, x ≥ 0, y ≥ 0.
Maximize Z = 6x + 4y, subject to x ≤ 2, x + y ≤ 3, -2x + y ≤ 1, x ≥ 0, y ≥ 0.
