Advertisements
Advertisements
प्रश्न
A gardener has supply of fertilizer of type I which consists of 10% nitrogen and 6% phosphoric acid and type II fertilizer which consists of 5% nitrogen and 10% phosphoric acid. After testing the soil conditions, he finds that he needs at least 14 kg of nitrogen and 14 kg of phosphoric acid for his crop. If the type I fertilizer costs 60 paise per kg and type II fertilizer costs 40 paise per kg, determine how many kilograms of each fertilizer should be used so that nutrient requirements are met at a minimum cost. What is the minimum cost?
Advertisements
उत्तर
Let x kg of type I fertilizer and y kg of type II fertilizer are supplied.
Quantity of fertilizer cannot be negative.
Therefore, \[x, y \geq 0\]
A gardener has supply of fertilizer of type I which consists of 10% nitrogen and type II fertilizer consists of 5% nitrogen and he needs at least 14 kg of nitrogen for his crop.\[\frac{10x}{100} + \frac{5x}{100} \geq 14 \Rightarrow 10x + 5x \geq 1400\]
A gardener has supply of fertilizer of type I which consists 6% phosphoric acid and type II fertilizer consists of 10% phosphoric acid.
And he needs 14 kg of phosphoric acid for his crop.
Therefore, according to the question, constraints are
\[10x + 5y \geq 1400\]
\[6x + 10y \geq 1400\]
Total cost = Z =\[0 . 60x + 0 . 40y\] which is to be minimised.
Min Z = \[0 . 60x + 0 . 40y\]subject to
\[10x + 5y \geq 1400\]
\[x, y \geq 0\]
6x + 10y = 1400, 10x + 5y = 1400, x = 0 and y = 0
Region represented by 6x + 10y ≥ 1400:
The line 6x + 10y = 1400 meets the coordinate axes at
Region represented by 10x + 5y ≥ 1400:
The line 10x + 5y = 1400 meets the coordinate axes at C(140, 0) and \[D\left( 0, 280 \right)\] respectively. By joining these points we obtain the line 10x + 5y = 1400. Clearly (0,0) does not satisfies the inequation 10x + 5y ≥ 1400. So,the region which does not contain the origin represents the solution set of the inequation 10x + 5y ≥ 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 6x + 10y ≥1400, 10x + 5y ≥ 1400, x ≥ 0 and y ≥ 0 are as follows.
| Corner point | Z= 0.60x + 0.40y |
| D | 112 |
| E | 92 |
| A | 140 |
The minimum value of Z is Rs 92 which is attained at E(100, 80).
Thus, the minimum cost is Rs 92 obtained when 100 kg of type I fertilizer and 80 kg of type II fertilizer were supplied.
APPEARS IN
संबंधित प्रश्न
Minimize `z=4x+5y ` subject to `2x+y>=7, 2x+3y<=15, x<=3,x>=0, y>=0` solve using graphical method.
Minimum and maximum z = 5x + 2y subject to the following constraints:
x-2y ≤ 2
3x+2y ≤ 12
-3x+2y ≤ 3
x ≥ 0,y ≥ 0
Find graphically, the maximum value of z = 2x + 5y, subject to constraints given below :
2x + 4y ≤ 83
x + y ≤ 6
x + y ≤ 4
x ≥ 0, y≥ 0
Solve the following L.P.P. graphically:
Minimise Z = 5x + 10y
Subject to x + 2y ≤ 120
Constraints x + y ≥ 60
x – 2y ≥ 0 and x, y ≥ 0
Solve the following LPP by graphical method:
Minimize Z = 7x + y subject to 5x + y ≥ 5, x + y ≥ 3, x ≥ 0, y ≥ 0
Maximise z = 8x + 9y subject to the constraints given below :
2x + 3y ≤ 6
3x − 2y ≤6
y ≤ 1
x, y ≥ 0
Minimize Z = 18x + 10y
Subject to
\[4x + y \geq 20\]
\[2x + 3y \geq 30\]
\[ x, y \geq 0\]
Maximize Z = 15x + 10y
Subject to
\[3x + 2y \leq 80\]
\[2x + 3y \leq 70\]
\[ x, y \geq 0\]
Minimize Z = 2x + 4y
Subject to
\[x + y \geq 8\]
\[x + 4y \geq 12\]
\[x \geq 3, y \geq 2\]
Maximize Z = 3x + 5y
Subject to
\[x + 2y \leq 20\]
\[x + y \leq 15\]
\[ y \leq 5\]
\[ x, y \geq 0\]
Find graphically, the maximum value of Z = 2x + 5y, subject to constraints given below:
2x + 4y ≤ 8
3x + y ≤ 6
x + y ≤ 4
x ≥ 0, 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 200 g of flour and 25 g of fat, and another kind of cake requires 100 g of flour and 50 g of fat. Find the maximum number of cakes which can be made from 5 kg of flour and 1 kg of fat assuming that there is no storage of the other ingredients used in making the cakes.
A dietician wishes to mix together two kinds of food X and Y in such a way that the mixture contains at least 10 units of vitamin A, 12 units of vitamin B and 8 units of vitamin C. The vitamin contents of one kg food is given below:
| Food | Vitamin A | Vitamin B | Vitamin C |
| X | 1 | 2 | 3 |
| Y | 2 | 2 | 1 |
One kg of food X costs ₹16 and one kg of food Y costs ₹20. Find the least cost of the mixture which will produce the required diet?
A firm manufactures headache pills in two sizes A and B. Size A contains 2 grains of aspirin, 5 grains of bicarbonate and 1 grain of codeine; size B contains 1 grain of aspirin, 8 grains of bicarbonate and 66 grains of codeine. It has been found by users that it requires at least 12 grains of aspirin, 7.4 grains of bicarbonate and 24 grains of codeine for providing immediate effects. Determine graphically the least number of pills a patient should have to get immediate relief. Determine also the quantity of codeine consumed by patient.
A small firm manufactures gold rings and chains. The total number of rings and chains manufactured per day is at most 24. It takes 1 hour to make a ring and 30 minutes to make a chain. The maximum number of hours available per day is 16. If the profit on a ring is Rs 300 and that on a chain is Rs 190, find the number of rings and chains that should be manufactured per day, so as to earn the maximum profit. Make it as an LPP and solve it graphically.
A merchant plans to sell two types of personal computers a desktop model and a portable model that will cost Rs 25,000 and Rs 40,000 respectively. He estimates that the total monthly demand of computers will not exceed 250 units. Determine the number of units of each type of computers which the merchant should stock to get maximum profit if he does not want to invest more than Rs 70 lakhs and his profit on the desktop model is Rs 4500 and on the portable model is Rs 5000. Make an LPP and solve it graphically.
A manufacturer has three machine I, II, III installed in his factory. Machines I and II are capable of being operated for at most 12 hours whereas machine III must be operated for atleast 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 each of M and N on the three machines are given in the following table:
| Items | Number of hours required on machines | ||
| I | II | III | |
| M | 1 | 2 | 1 |
| N | 2 | 1 | 1.25 |
She makes a profit of ₹600 and ₹400 on items M and N respectively. How many of each item should she produce so as to maximise her profit assuming that she can sell all the items that she produced? What will be the maximum profit?
A manufacturer considers that men and women workers are equally efficient and so he pays them at the same rate. He has 30 and 17 units of workers (male and female) and capital respectively, which he uses to produce two types of goods A and B. To produce one unit of A, 2 workers and 3 units of capital are required while 3 workers and 1 unit of capital is required to produce one unit of B. If A and B are priced at ₹100 and ₹120 per unit respectively, how should he use his resources to maximise the total revenue? Form the above as an LPP and solve graphically. Do you agree with this view of the manufacturer that men and women workers are equally efficient and so should be paid at the same rate?
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
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.
A medical company has factories at two places, A and B. From these places, supply is made to each of its three agencies situated at P, Q and R. The monthly requirements of the agencies are respectively 40, 40 and 50 packets of the medicines, while the production capacity of the factories, A and B, are 60 and 70 packets respectively. The transportation cost per packet from the factories to the agencies are given below:
| Transportation Cost per packet(in Rs.) | ||
| From-> | A | B |
| To | ||
| P | 5 | 4 |
| Q | 4 | 2 |
| R | 3 | 5 |
The point at which the maximum value of x + y subject to the constraints x + 2y ≤ 70, 2x + y ≤ 95, x ≥ 0, y ≥ 0 is obtained, is ______.
A carpenter has 90, 80 and 50 running feet respectively of teak wood, plywood and rosewood which is used to product A and product B. Each unit of product A requires 2, 1 and 1 running feet and each unit of product B requires 1, 2 and 1 running feet of teak wood, plywood and rosewood respectively. If product A is sold for Rs. 48 per unit and product B is sold for Rs. 40 per unit, how many units of product A and product B should be produced and sold by the carpenter, in order to obtain the maximum gross income? Formulate the above as a Linear Programming Problem and solve it, indicating clearly the feasible region in the graph.
A company manufactures two types of novelty souvenirs made of plywood. Souvenirs of type A
require 5 minutes each for cutting and 10 minutes each for assembling. Souvenirs of type B require 8 minutes each for cutting and 8 minutes each for assembling. There are 3 hours and 20 minutes available for cutting and 4 hours available for assembling. The profit is Rs. 50 each for type A and Rs. 60 each for type B souvenirs. How many souvenirs of each type should the company manufacture in order to maximize profit? Formulate the above LPP and solve it graphically and also find the maximum profit.
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.
Sketch the graph of inequation x ≥ 5y in xoy co-ordinate system
Let R be the feasible region (convex polygon) for a linear programming problem and let Z = ax + by be the objective function. When Z has an optimal value (maximum or minimum), where the variables x and y are subject to constraints described by linear inequalities,
The feasible region (shaded) for a L.P.P is shown in the figure. The maximum Z = 5x + 7y 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 objective function Z = x1 + x2, subject to the constraints are x1 + x2 ≤ 10, – 2x1 + 3x2 ≤ 15, x1 ≤ 6, x1, x2 ≥ 0, has maximum value ______ of the feasible region.
Solve the following Linear Programming Problem graphically.
Maximise Z = 5x + 2y subject to:
x – 2y ≤ 2,
3x + 2y ≤ 12,
– 3x + 2y ≤ 3,
x ≥ 0, y ≥ 0
Find feasible solution for the following system of linear inequation graphically.
3x + 4y ≥ 12, 4x + 7y ≤ 28, x ≥ 0, y ≥ 0
Which description represents an unbounded region?
For \[x+y\leq50\], \[3x+y\leq90\], \[x\geq0\], and \[y\geq0\], which is a corner point of the feasible region?
For \[Z=4x+y\], which corner point gives the maximum value subject to \[x+y\leq50\], \[3x+y\leq90\], \[x\geq0\], and \[y\geq0\]?
If two corner points give the same optimum value, which points are also optimal?
