Advertisements
Advertisements
Question
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?
Solution
Let x large boxes and y small boxes be manufactured.
Number of boxes cannot be negative.
Therefore, \[x \geq 0, y \geq 0\]
The large boxes require 4 sq. metre per box while the small boxes require 3 sq. metre per box and if 60 sq. metre of cardboard is in stock.
\[4x + 3y \leq 60\]
The manufacturer is required to make at least three large boxes and at least twice as many small boxes as large boxes.
\[x \geq 3\]
\[y \geq 2x\]
If the profits on the large and small boxes are Rs 3 and Rs 2 per box. Therefore, profit gained by him on x large boxes and y small boxes is Rs 3x and Rs 2y respectively.
Total profit = Z = \[3x + 2y\]
The mathematical formulation of the given problem is
Max Z = \[3x + 2y\]
subject to
\[4x + 3y \leq 60\]
\[x \geq 3\]
\[y \geq 2x\]
\[x \geq 0, y \geq 0\]
First we will convert inequations into equations as follows:
4x + 3y = 60, x = 3, y = 2x, x = 0 and y = 0
Region represented by 4x + 3y ≤ 60:
The line 4x + 3y = 60 meets the coordinate axes at A(15, 0) and B(0, 20) respectively. By joining these points we obtain the line
4x + 3y = 60. Clearly (0,0) satisfies the 4x + 3y = 60. So, the region which contains the origin represents the solution set of the inequation 4x + 3y ≤ 60.
Region represented by x ≥ 3:
The line x = 3 is the line passes through (3, 0) and is parallel to Y axis. The region to the right of the line x = 3 will satisfy the inequation
x ≥ 3.
Region represented by y ≥ 2x:
The line y = 2x is the line that passes through (0, 0). The region above the line y = 2x will satisfy the inequation y ≥ 2x. Like if we take an example taking a point (5, 1) below the line y = 2x .Here, 1 < 10 which does not satisfies the inequation y ≥ 2x. Hence, the region above the line y = 2x will satisfy the inequation y ≥ 2x.
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 4x + 3y ≤ 60, x ≥ 3, y ≥ 2x, x ≥ 0 and y ≥ 0 are as follows
The corner points are E \[\left( 3, 16 \right)\], D(6, 12) and C(3, 6).The values of Z at the corner points are
| Corner points | Z = \[3x + 2y\] |
| E | 41 |
| D | 42 |
| C | 21 |
APPEARS IN
RELATED QUESTIONS
Solve the following L.P.P graphically:
Maximize: Z = 10x + 25y
Subject to: x ≤ 3, y ≤ 3, x + y ≤ 5, x ≥ 0, y ≥ 0
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
Maximise Z = x + 2y subject to the constraints
`x + 2y >= 100`
`2x - y <= 0`
`2x + y <= 200`
Solve the above LPP graphically
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
Solve the following L.P.P graphically: Maximise Z = 20x + 10y
Subject to the following constraints x + 2y ≤ 28,
3x + y ≤ 24,
x ≥ 2,
x, y ≥ 0
Solve the following LPP graphically :
Maximise Z = 105x + 90y
subject to the constraints
x + y ≤ 50
2x + y ≤ 80
x ≥ 0, y ≥ 0.
Maximize Z = 5x + 3y
Subject to
\[3x + 5y \leq 15\]
\[5x + 2y \leq 10\]
\[ x, y \geq 0\]
Maximize Z = 15x + 10y
Subject to
\[3x + 2y \leq 80\]
\[2x + 3y \leq 70\]
\[ x, y \geq 0\]
Maximize Z = 7x + 10y
Subject to
\[x + y \leq 30000\]
\[ y \leq 12000\]
\[ x \geq 6000\]
\[ x \geq y\]
\[ x, y \geq 0\]
Maximize Z = 4x + 3y
Subject to
\[3x + 4y \leq 24\]
\[8x + 6y \leq 48\]
\[ x \leq 5\]
\[ y \leq 6\]
\[ x, y \geq 0\]
Maximize Z = 3x1 + 4x2, if possible,
Subject to the constraints
\[x_1 - x_2 \leq - 1\]
\[ - x_1 + x_2 \leq 0\]
\[ x_1 , x_2 \geq 0\]
Solve the following linear programming problem graphically:
Minimize z = 6 x + 3 y
Subject to the constraints:
4 x + \[y \geq\] 80
x + 5 \[y \geq\] 115
3 x + 2 \[y \leq\] 150
\[x \geq\] 0 , \[y \geq\] 0
A hospital dietician wishes to find the cheapest combination of two foods, A and B, that contains at least 0.5 milligram of thiamin and at least 600 calories. Each unit of Acontains 0.12 milligram of thiamin and 100 calories, while each unit of B contains 0.10 milligram of thiamin and 150 calories. If each food costs 10 paise per unit, how many units of each should be combined at a minimum cost?
Kellogg is a new cereal formed of a mixture of bran and rice that contains at least 88 grams of protein and at least 36 milligrams of iron. Knowing that bran contains 80 grams of protein and 40 milligrams of iron per kilogram, and that rice contains 100 grams of protein and 30 milligrams of iron per kilogram, find the minimum cost of producing this new cereal if bran costs Rs 5 per kg and rice costs Rs 4 per kg
A wholesale dealer deals in two kinds, A and B (say) of mixture of nuts. Each kg of mixture A contains 60 grams of almonds, 30 grams of cashew nuts and 30 grams of hazel nuts. Each kg of mixture B contains 30 grams of almonds, 60 grams of cashew nuts and 180 grams of hazel nuts. The remainder of both mixtures is per nuts. The dealer is contemplating to use mixtures A and B to make a bag which will contain at least 240 grams of almonds, 300 grams of cashew nuts and 540 grams of hazel nuts. Mixture A costs Rs 8 per kg. and mixture B costs Rs 12 per kg. Assuming that mixtures A and B are uniform, use graphical method to determine the number of kg. of each mixture which he should use to minimise the cost of the bag.
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?
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 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 company manufactures two articles A and B. There are two departments through which these articles are processed: (i) assembly and (ii) finishing departments. The maximum capacity of the first department is 60 hours a week and that of other department is 48 hours per week. The product of each unit of article A requires 4 hours in assembly and 2 hours in finishing and that of each unit of B requires 2 hours in assembly and 4 hours in finishing. If the profit is Rs 6 for each unit of A and Rs 8 for each unit of B, find the number of units of A and B to be produced per week in order to have maximum profit.
If a young man drives his vehicle at 25 km/hr, he has to spend ₹2 per km on petrol. If he drives it at a faster speed of 40 km/hr, the petrol cost increases to ₹5 per km. He has ₹100 to spend on petrol and travel within one hour. Express this as an LPP and solve the same.
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?
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 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.
Find the solution set of inequalities 0 ≤ x ≤ 5, 0 ≤ 2y ≤ 7
The maximum value of Z = 5x + 4y, Subject to y ≤ 2x, x ≤ 2y, x + y ≤ 3, x ≥ 0, y ≥ 0 is ______.
If 4x + 5y ≤ 20, x + y ≥ 3, x ≥ 0, y ≥ 0, maximum 2x + 3y is ______.
The minimum value of z = 7x + 9y subject to 3x + y ≤ 6, 5x + 8y ≤ 40, x ≥ 0, y ≥ 2 is ______.
Of all the points of the feasible region for maximum or minimum of objective function the points.
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,
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 ____________.
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 2x + y subject to 3x + 5y ≤ 26 and 5x + 3y ≤ 30, x ≥ 0, y ≥ 0 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.
The objective function Z = ax + by of an LPP has maximum vaiue 42 at (4, 6) and minimum value 19 at (3, 2). Which of the following is true?
Solve the following Linear Programming Problem graphically:
Minimize: z = x + 2y,
Subject to the constraints: x + 2y ≥ 100, 2x – y ≤ 0, 2x + y ≤ 200, x, y ≥ 0.
Aman has ₹ 1500 to purchase rice and wheat for his grocery shop. Each sack of rice and wheat costs ₹ 180 and Rupee ₹ 120 respectively. He can store a maximum number of 10 bags in his shop. He will earn a profit of ₹ 11 per bag of rice and ₹ 9 per bag of wheat.
- Formulate a Linear Programming Problem to maximise Aman’s profit.
- Calculate the maximum profit.
