Advertisements
Advertisements
प्रश्न
A diet for a sick person must contain at least 4000 units of vitamins, 50 units of minerals and 1400 of calories. Two foods A and B, are available at a cost of Rs 4 and Rs 3 per unit respectively. If one unit of A contains 200 units of vitamin, 1 unit of mineral and 40 calories and one unit of food B contains 100 units of vitamin, 2 units of minerals and 40 calories, find what combination of foods should be used to have the least cost?
Advertisements
उत्तर
Let the sick person takes x units and y units of food I and II respectively that were taken in the diet.
Since, per unit of food I costs Rs 4 and that of food II costs Rs 3.
Therefore, x units of food I costs Rs 4x and y units of food II costs Rs 3y.
Total cost = Rs (4x + 3y)
Let Z denote the total cost
Then, Z = 4x + 3y
If one unit of A contains 200 units of vitamin and one unit of food B contains 100 units of vitamin.
Thus, x units of food I and y units of food II contains 200x + 100y units of vitamin.
But a diet for a sick person must contain at least 4000 units of vitamins.
∴ \[200x + 100y \geq 4000\]
If one unit of A contains 1 unit of mineral and one unit of food B contains 2 units of mineral.
Thus, x units of food I and y units of food II contains x + 2y units of mineral.
But a diet for a sick person must contain at least 50 units of vitamins.
∴ \[x + 2y \geq 50\] If one unit of A contains 40 calories and one unit of food B contains 40 calories.
Thus, x units of food I and y units of food II contains 40x + 40y units of calories.
But a diet for a sick person must contain at least 1400 calories.
So
Min Z = 4x + 3y
subject to
\[x + 2y \geq 50\]
\[40x + 40y \geq 1400\]
\[x, y \geq 0\]
First, we will convert the given inequations into equations, we obtain the following equations:
200x + 100y = 4000, x +2y = 50, 40x + 40y =1400, x = 0 and y = 0
Region represented by 200x + 100y ≥ 4000:
The line 200x + 100y = 4000 meets the coordinate axes at A1(20, 0) and B1(0,40) respectively. By joining these points we obtain the line
200x + 100y = 4000.Clearly (0,0) does not satisfies the inequation 200x + 100y ≥ 4000. So,the region in xy plane which does not contain the origin represents the solution set of the inequation 200x + 100y ≥ 4000.
Region represented by x +2y ≥ 50:
The line x +2y = 50 meets the coordinate axes at C1(50, 0) and D1(0, 25) respectively. By joining these points we obtain the line
x +2y = 50.Clearly (0,0) does not satisfies the x +2y ≥ 50. So,the region which does not contains the origin represents the solution set of the inequation x +2y ≥ 50.
Region represented by 40x + 40y ≥ 1400:
The line 40x + 40y = 1400 meets the coordinate axes at E1(35, 0) and F1(0, 35) respectively. By joining these points we obtain the line
40x + 40y = 1400.Clearly (0,0) does not satisfies the inequation 40x + 40y ≥ 1400. So,the region which does not contains the origin represents the solution set of the inequation 40x + 40y ≥ 1400.
Region represented by x ≥ 0 and 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 200x + 100y ≥ 4000,x +2y ≥ 50, 40x + 40y ≥ 1400, x ≥ 0, and y ≥ 0 are as follows.
The corner points of the feasible region are B1(0, 40), G1(5, 30), H1(20, 15) and C1(50, 0)
The value of the objective function at these points are given by the following table
| Points | Value of Z |
| B1 | 4(0)+3(40) = 120 |
| G1 | 4(5)+3(30) = 110 |
| H1 | 4(20) + 3(15) = 125 |
| C1 | 4(50)+3(0) = 200 |
The minimum cost is Rs 110 which is at G1(5, 30).
Hence, the required combination of food is 5 units of food A and 30 units of food B.
APPEARS IN
संबंधित प्रश्न
Minimize: Z = 6x + 4y
Subject to the conditions:
3x + 2y ≥ 12,
x + y ≥ 5,
0 ≤ x ≤ 4,
0 ≤ y ≤ 4
There are two types of fertilisers 'A' and 'B'. 'A' consists of 12% nitrogen and 5% phosphoric acid whereas 'B' consists of 4% nitrogen and 5% phosphoric acid. After testing the soil conditions, farmer finds that he needs at least 12 kg of nitrogen and 12 kg of phosphoric acid for his crops. If 'A' costs Rs 10 per kg and 'B' cost Rs 8 per kg, then graphically determine how much of each type of fertiliser should be used so that nutrient requirements are met at a minimum cost
A manufacturer produces two products A and B. Both the products are processed on two different machines. The available capacity of first machine is 12 hours and that of second machine is 9 hours per day. Each unit of product A requires 3 hours on both machines and each unit of product B requires 2 hours on first machine and 1 hour on second machine. Each unit of product A is sold at Rs 7 profit and B at a profit of Rs 4. Find the production level per day for maximum profit graphically.
A manufacturing company makes two types of teaching aids A and B of Mathematics for class XII. Each type of A requires 9 labour hours for fabricating and 1 labour hour for finishing. Each type of B requires 12 labour hours for fabricating and 3 labour hours for finishing. For fabricating and finishing, the maximum labour hours available per week are 180 and 30, respectively. The company makes a profit of Rs 80 on each piece of type A and Rs 120 on each piece of type B. How many pieces of type A and type B should be manufactured per week to get maximum profit? Make it as an LPP and solve graphically. What is the maximum profit per week?
Solve the following LPP by graphical method:
Maximize: z = 3x + 5y
Subject to: x + 4y ≤ 24
3x + y ≤ 21
x + y ≤ 9
x ≥ 0, y ≥ 0
Also find the maximum value of z.
Maximize Z = 7x + 10y
Subject to
\[x + y \leq 30000\]
\[ y \leq 12000\]
\[ x \geq 6000\]
\[ x \geq y\]
\[ x, y \geq 0\]
Minimize Z = 2x + 4y
Subject to
\[x + y \geq 8\]
\[x + 4y \geq 12\]
\[x \geq 3, y \geq 2\]
Maximize Z = x + y
Subject to
\[- 2x + y \leq 1\]
\[ x \leq 2\]
\[ x + y \leq 3\]
\[ x, y \geq 0\]
Find the maximum and minimum value of 2x + y subject to the constraints:
x + 3y ≥ 6, x − 3y ≤ 3, 3x + 4y ≤ 24, − 3x + 2y ≤ 6, 5x + y ≥ 5, x, y ≥ 0.
Find the minimum value of 3x + 5y subject to the constraints
− 2x + y ≤ 4, x + y ≥ 3, x − 2y ≤ 2, x, y ≥ 0.
A diet is to contain at least 80 units of vitamin A and 100 units of minerals. Two foods F1and F2 are available. Food F1 costs Rs 4 per unit and F2 costs Rs 6 per unit one unit of food F1 contains 3 units of vitamin A and 4 units of minerals. One unit of food F2contains 6 units of vitamin A and 3 units of minerals. Formulate this as a linear programming problem and find graphically the minimum cost for diet that consists of mixture of these foods and also meets the mineral nutritional requirements
One kind of cake requires 300 gm of flour and 15 gm of fat, another kind of cake requires 150 gm of flour and 30 gm of fat. Find the maximum number of cakes which can be made from 7.5 kg of flour and 600 gm of fat, assuming that there is no shortage of the other ingradients used in making the cake. Make it as an LPP and solve it graphically.
If a young man drives his vehicle at 25 km/hr, he has to spend Rs 2 per km on petrol. If he drives it at a faster speed of 40 km/hr, the petrol cost increases to Rs 5/per km. He has Rs 100 to spend on petrol and travel within one hour. Express this as an LPP and solve the same.
Two tailors, A and B earn Rs 15 and Rs 20 per day respectively. A can stitch 6 shirts and 4 pants while B can stitch 10 shirts and 4 pants per day. How many days shall each work if it is desired to produce (at least) 60 shirts and 32 pants at a minimum labour cost?
A firm manufacturing two types of electric items, A and B, can make a profit of Rs 20 per unit of A and Rs 30 per unit of B. Each unit of A requires 3 motors and 4 transformers and each unit of B requires 2 motors and 4 transformers. The total supply of these per month is restricted to 210 motors and 300 transformers. Type B is an export model requiring a voltage stabilizer which has a supply restricted to 65 units per month. Formulate the linear programing problem for maximum profit and solve it graphically.
A manufacturer produces two types of steel trunks. He has two machines A and B. For completing, the first types of the trunk requires 3 hours on machine A and 3 hours on machine B, whereas the second type of the trunk requires 3 hours on machine A and 2 hours on machine B. Machines A and B can work at most for 18 hours and 15 hours per day respectively. He earns a profit of Rs 30 and Rs 25 per trunk of the first type and the second type respectively. How many trunks of each type must he make each day to make maximum profit?
Anil wants to invest at most Rs 12000 in Saving Certificates and National Saving Bonds. According to rules, he has to invest at least Rs 2000 in Saving Certificates and at least Rs 4000 in National Saving Bonds. If the rate of interest on saving certificate is 8% per annum and the rate of interest on National Saving Bond is 10% per annum, how much money should he invest to earn maximum yearly income? Find also his maximum yearly income.
A firm manufactures two types of products A and B and sells them at a profit of Rs 5 per unit of type A and Rs 3 per unit of type B. Each product is processed on two machines M1 and M2. One unit of type A requires one minute of processing time on M1 and two minutes of processing time on M2, whereas one unit of type B requires one minute of processing time on M1 and one minute on M2. Machines M1 and M2 are respectively available for at most 5 hours and 6 hours in a day. Find out how many units of each type of product should the firm produce a day in order to maximize the profit. Solve the problem graphically.
A factory makes tennis rackets and cricket bats. A tennis racket takes 1.5 hours of machine time and 3 hours of craftman's time in its making while a cricket bat takes 3 hours of machine time and 1 hour of craftman's time. In a day, the factory has the availability of not more than 42 hours of machine time and 24 hours of craftman's time.
(i) What number of rackets and bats must be made if the factory is to work at full capacity?
(ii) If the profit on a racket and on a bat is Rs 20 and Rs 10 respectively, find the maximum profit of the factory when it works at full capacity.
A manufacturer produces two products A and B. Both the products are processed on two different machines. The available capacity of first machine is 12 hours and that of second machine is 9 hours per day. Each unit of product A requires 3 hours on both machines and each unit of product B requires 2 hours on first machine and 1 hour on second machine. Each unit of product A is sold at ₹7 profit and that of B at a profit of ₹4. Find the production level per day for maximum profit graphically.
There are two types of fertilisers 'A' and 'B' . 'A' consists of 12% nitrogen and 5% phosphoric acid whereas 'B' consists of 4% nitrogen and 5% phosphoric acid. After testing the soil conditions, farmer finds that he needs at least 12 kg of nitrogen and 12 kg of phosphoric acid for his crops. If 'A' costs ₹10 per kg and 'B' cost ₹8 per kg, then graphically determine how much of each type of fertiliser should be used so that nutrient requiremnets are met at a minimum cost
By graphical method, the solution of linear programming problem
\[\text{ Subject } to \text{ 3 } x_1 + 2 x_2 \leq 18\]
\[ x_1 \leq 4\]
\[ x_2 \leq 6\]
\[ x_1 \geq 0, x_2 \geq 0, \text{ is } \]
Maximize: z = 3x + 5y Subject to
x +4y ≤ 24 3x + y ≤ 21
x + y ≤ 9 x ≥ 0 , y ≥0
A farmer has a supply of chemical fertilizer of type A which contains 10% nitrogen and 6% phosphoric acid and of type B which contains 5% nitrogen and 10% phosphoric acid. After the soil test, it is found that at least 7 kg of nitrogen and the same quantity of phosphoric acid is required for a good crop. The fertilizer of type A costs ₹ 5.00 per kg and the type B costs ₹ 8.00 per kg. Using Linear programming, find how many kilograms of each type of fertilizer should be bought to meet the requirement and for the cost to be minimum. Find the feasible region in the graph.
A company manufactures two types of products A and B. Each unit of A requires 3 grams of nickel and 1 gram of chromium, while each unit of B requires 1 gram of nickel and 2 grams of chromium. The firm can produce 9 grams of nickel and 8 grams of chromium. The profit is ₹ 40 on each unit of the product of type A and ₹ 50 on each unit of type B. How many units of each type should the company manufacture so as to earn a maximum profit? Use linear programming to find the solution.
A company manufactures two types of cardigans: type A and type B. It costs ₹ 360 to make a type A cardigan and ₹ 120 to make a type B cardigan. The company can make at most 300 cardigans and spend at most ₹ 72000 a day. The number of cardigans of type B cannot exceed the number of cardigans of type A by more than 200. The company makes a profit of ₹ 100 for each cardigan of type A and ₹ 50 for every cardigan of type B.
Formulate this problem as a linear programming problem to maximize the profit to the company. Solve it graphically and find the maximum profit.
Find the graphical solution for the system of linear inequation 2x + y ≤ 2, x − y ≤ 1
Draw the graph of inequalities x ≤ 6, y −2 ≤ 0, x ≥ 0, y ≥ 0 and indicate the feasible region
The constraints of an LPP are 7 ≤ x ≤ 12, 8 ≤ y ≤ 13. Determine the vertices of the feasible region formed by them.
If 4x + 5y ≤ 20, x + y ≥ 3, x ≥ 0, y ≥ 0, maximum 2x + 3y is ______.
Z = 20x1 + 20x2, subject to x1 ≥ 0, x2 ≥ 0, x1 + 2x2 ≥ 8, 3x1 + 2x2 ≥ 15, 5x1 + 2x2 ≥ 20. The minimum value of Z occurs at ____________.
In the Corner point method for solving a linear programming problem the second step after finding the feasible region of the linear programming problem and determining its corner points is ____________.
Which of the statements describe the solution set for `-2(x + 8) = - 2x + 20`?
The comer point of the feasible region determined by the following system of linear inequalities:
2x + y ≤ 10, x + 3y ≤ 15, x, y ≥ 0 are (0, 0), (5, 0), (3, 4) and (0, 5). Let x = Px + qx where P, q > 0 condition on P and Q so that the maximum of z occurs at both (3, 4) and (0, 5) is
The corner points of the shaded unbounded feasible region of an LPP are (0, 4), (0.6, 1.6) and (3, 0) as shown in the figure. The minimum value of the objective function Z = 4x + 6y occurs at ______.
The feasible region corresponding to the linear constraints of a Linear Programming Problem is given below.
Which of the following is not a constraint to the given Linear Programming Problem?
If x – y ≥ 8, x ≥ 3, y ≥ 3, x ≥ 0, y ≥ 0 then find the coordinates of the corner points of the feasible region.
