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
Solve the following LPP by using graphical method.
Maximize : Z = 6x + 4y
Subject to x ≤ 2, x + y ≤ 3, -2x + y ≤ 1, x ≥ 0, y ≥ 0.
Also find maximum value of Z.
Solve the following L.P.P graphically:
Maximize: Z = 10x + 25y
Subject to: x ≤ 3, y ≤ 3, x + y ≤ 5, x ≥ 0, y ≥ 0
Solve the following L.P.P. graphically Maximise Z = 4x + y
Subject to following constraints x + y ≤ 50
3x + y ≤ 90,
x ≥ 10
x, y ≥ 0
Maximise z = 8x + 9y subject to the constraints given below :
2x + 3y ≤ 6
3x − 2y ≤6
y ≤ 1
x, y ≥ 0
Maximize Z = x + y
Subject to
\[- 2x + y \leq 1\]
\[ x \leq 2\]
\[ x + y \leq 3\]
\[ x, y \geq 0\]
Maximize Z = 3x + 3y, if possible,
Subject to the constraints
\[x - y \leq 1\]
\[x + y \geq 3\]
\[ x, y \geq 0\]
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 of two foods F1 and F2 contains nutrients thiamine, phosphorous and iron. The amount of each nutrient in each of the food (in milligrams per 25 gms) is given in the following table:
Nutrients |
Food |
F1 | F2 |
| Thiamine | 0.25 | 0.10 |
|
| Phosphorous | 0.75 | 1.50 | |
| Iron | 1.60 | 0.80 | |
The minimum requirement of the nutrients in the diet are 1.00 mg of thiamine, 7.50 mg of phosphorous and 10.00 mg of iron. The cost of F1 is 20 paise per 25 gms while the cost of F2 is 15 paise per 25 gms. Find the minimum cost of diet.
A company produces two types of leather belts, say type A and B. Belt A is a superior quality and belt B is of a lower quality. Profits on each type of belt are Rs 2 and Rs 1.50 per belt, respectively. Each belt of type A requires twice as much time as required by a belt of type B. If all belts were of type B, the company could produce 1000 belts per day. But the supply of leather is sufficient only for 800 belts per day (both A and B combined). Belt A requires a fancy buckle and only 400 fancy buckles are available for this per day. For belt of type B, only 700 buckles are available per day.
How should the company manufacture the two types of belts in order to have a maximum overall profit?
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:
| Machines | |||
| I | II | III | |
| A B |
12 6 |
18 0 |
6 9 |
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.
A factory uses three different resources for the manufacture of two different products, 20 units of the resources A, 12 units of B and 16 units of C being available. 1 unit of the first product requires 2, 2 and 4 units of the respective resources and 1 unit of the second product requires 4, 2 and 0 units of respective resources. It is known that the first product gives a profit of 2 monetary units per unit and the second 3. Formulate the linear programming problem. How many units of each product should be manufactured for maximizing the profit? Solve it graphically.
A chemical company produces two compounds, A and B. The following table gives the units of ingredients, C and D per kg of compounds A and B as well as minimum requirements of C and D and costs per kg of A and B. Find the quantities of A and B which would give a supply of C and D at a minimum cost.
| Compound | Minimum requirement | ||
| A | B | ||
| Ingredient C Ingredient D |
1 3 |
2 1 |
80 75 |
| Cost (in Rs) per kg | 4 | 6 | - |
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 20 minutes available for cutting and 4 hours available for assembling. The profit is 50 paise each for type A and 60 paise each for type B souvenirs. How many souvenirs of each type should the company manufacture in order to maximize the profit?
A box manufacturer makes large and small boxes from a large piece of cardboard. The large boxes require 4 sq. metre per box while the small boxes require 3 sq. metre per box. The manufacturer is required to make at least three large boxes and at least twice as many small boxes as large boxes. If 60 sq. metre of cardboard is in stock, and if the profits on the large and small boxes are Rs 3 and Rs 2 per box, how many of each should be made in order to maximize the total profit?
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 cooperative society of farmers has 50 hectares of land to grow two crops X and Y. The profits from crops X and Y per hectare are estimated as ₹10,500 and ₹9,000 respectively. To control weeds, a liquid herbicide has to be used for crops X and Y at the rate of 20 litres and 10 litres per hectare, respectively. Further not more than 800 litres of herbicide should be used in order to protect fish and wildlife using a pond which collects drainage from this land. How much land should be allocated to each crop so as to maximise the total profit of the society?
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 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 |
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 manufacturer has employed 5 skilled men and 10 semi-skilled men and makes two models A and B of an article. The making of one item of model A requires 2 hours of work by a skilled man and 2 hours work by a semi-skilled man. One item of model B requires 1 hour by a skilled man and 3 hours by a semi-skilled man. No man is expected to work more than 8 hours per day. The manufacturer's profit on an item of model A is ₹ 15 and on an item of model B is ₹ 10. How many items of each model should be made per day in order to maximize daily profit? Formulate the above LPP and solve it graphically and find the maximum profit.
Sketch the graph of inequation x ≥ 5y in xoy co-ordinate system
Maximum value of 4x + 13y subject to constraints x ≥ 0, y ≥ 0, x + y ≤ 5 and 3x + y ≤ 9 is ______.
Area of the region bounded by y = cos x, x = 0, x = π and X-axis is ______ sq.units.
The feasible region of an LPP is shown in the figure. If z = 3x + 9y, then the minimum value of z occurs at ______.
For the LPP, maximize z = x + 4y subject to the constraints x + 2y ≤ 2, x + 2y ≥ 8, x, y ≥ 0 ______.
In linear programming feasible region (or solution region) for the problem is ____________.
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,
Any point in the feasible region that gives the optional value (maximum or minimum) of the objective function is called:-
Solve the following Linear Programming Problem graphically:
Maximize Z = 400x + 300y subject to x + y ≤ 200, x ≤ 40, x ≥ 20, y ≥ 0
The maximum value of z = 5x + 2y, subject to the constraints x + y ≤ 7, x + 2y ≤ 10, x, y ≥ 0 is ______.
The maximum value of 2x + y subject to 3x + 5y ≤ 26 and 5x + 3y ≤ 30, x ≥ 0, y ≥ 0 is ______.
Solve the following Linear Programming Problem graphically:
Maximize: P = 70x + 40y
Subject to: 3x + 2y ≤ 9,
3x + y ≤ 9,
x ≥ 0,y ≥ 0.
Minimize z = x + 2y,
Subject to x + 2y ≥ 50, 2x – y ≤ 0, 2x + y ≤ 100, x ≥ 0, y ≥ 0.
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
A linear programming problem is given by Z = px + qy, where p, q > 0 subject to the constraints x + y ≤ 60, 5x + y ≤ 100, x ≥ 0 and y ≥ 0.
- Solve graphically to find the corner points of the feasible region.
- If Z = px + qy is maximum at (0, 60) and (10, 50), find the relation of p and q. Also mention the number of optimal solution(s) in this case.
What is called the feasible region?
Which description represents a bounded region?
For \[x+y\leq50\], \[3x+y\leq90\], \[x\geq0\], and \[y\geq0\], which is a corner point of the feasible region?
What is the value of \[Z=4x+y\] at the corner point \[(30,0)\]?
