Advertisements
Advertisements
Question
A carpenter makes chairs and tables, profits are ₹ 140 per chair and ₹ 210 per table. Both products are processed on three machines, Assembling, Finishing and Polishing. The time required for each product in hours and the availability of each machine is given by the following table.
| Product/Machines | Chair (x) |
Table (y) |
Available time (hours) |
| Assembling | 3 | 3 | 36 |
| Finishing | 5 | 2 | 50 |
| Polishing | 2 | 6 | 60 |
Formulate and solve the following Linear programming problems using graphical method.
Advertisements
Solution
Let x be the number of chairs and y be the number of tables.
∴ The constraints are
3x + 3y ≤ 36
5x + 2y ≤ 50
2x + 6y ≤ 60
Since x and y are the numbers of chairs and tables, respectively.
∴ They cannot be negative.
∴ x ≥ 0, y ≥ 0
Now, the profit for one chair is ₹ 140 and the profit for one table is ₹ 210
∴ Total profit (Z) = 140x + 210y
This is an objective function to be maximized
∴ The given problem can be formulated as
Maximize Z = 140x + 210y
Subject to 3x + 3y ≤ 36
5x + 2y ≤ 50
2x + 6y ≤ 60
x ≥ 0, y ≥ 0
To find the graphical solution, construct the table as follows:
| Inequation | Equation | Double intercept form | Points (x, y) | Region |
| 3x + 3y ≤ 36 | 3x + 3y = 36 | `x/(2) + y/(12)` = 1 | A (12, 0) B (0, 12) |
3(0) + 3(0) ≤ 36 ∴ (0) ≤ 36 ∴ Origin-side |
| 5x + 2y ≤ 50 | 5x + 2y = 50 | `x/(10) + y/(25)` = 1 | C 10, 0) D (0, 25) |
5(0) + 2(0) ≤ 50 ∴ 0 ≤ 50 ∴ Origin-side |
| 2x + 6y ≤60 | 2x + 6y = 60 | `x/(30) + y/(10)` = 1 | E (30, 0) F (0, 10 |
2(0) + 6(0) ≤ 60 ∴ 0 ≤ 60 ∴ Origin-side |
| x ≥ 0 | x = 0 | – | – | R.H.S. of Y-axis |
| y ≥ 0 | y = 0 | – | – | above X-axis |
The shaded portion OFG HC is the feasible region,
Whose vertices are O (0, 0), F (0, 10), G, H and C (10, 0)
G is the point of intersection of lines.
2x + 6y = 60
i.e., x + 3y = 30 …(i)
and 3x + 3y = 36
i.e., x + y = 12 …(ii)
∴ By (i) – (ii), we get
x + 3y = 30
x + y = 12
– – –
2y = 18
∴ y = 9
Substituting y = 9 in (ii), we get
x + 9 = 12
∴ x = 12 – 9
∴ x = 3
∴ G = (3, 9)
H is the point of intersection of lines.
3x + 3y = 36
i.e., x + y = 12 …(ii)
5x + 2y = 50 …(iii)
∴ By 2 x (ii) – (iii), we get
2x + 2y = 24
5x + 2y = 50
– – –
– 3x – 26
∴ x = `(26)/(3)`
Substituting x = `(26)/(3)` in (ii), we get
`(26)/(3) + y` = 12
∴ y = `12 - (26)/(3) = (36 - 26)/(3)`
∴ y = `(10)/(3)`
∴ H`(26/3, 10/3)`
Here, the objective function is Z = 140x + 210y
Now, we will find the maximum value of Z as follows:
| Feasible Points | The value of Z = 140x + 210y |
| O (0, 0) | Z = 140(0) + 210(0) = 0 |
| F (0, 10) | Z = 140(0) + 210(10) = 2100 |
| G (3, 9) | Z = 140(3) + 210(9) = 420 + 1890 = 2310 |
| H`(36/3, 10/3)` | Z = `140(26/3) + 210(10/3) = (3640)/(3) + (2100)/(3)` = 1913.33 |
| C (10, 0) | Z = 140(10) + 210(0) = 1400 |
∴ Z has a maximum value of 2310 at G (3, 9)
∴ Maximum profit is ₹ 2310, when x = number of chairs = 3, y = number of tables = 9.
APPEARS IN
RELATED QUESTIONS
A company produces two types of goods A and B, that require gold and silver. Each unit of type A requires 3 g of silver and 1 g of golds while that of type B requires 1 g of silver and 2 g of gold. The company can procure a maximum of 9 g of silver and 8 g of gold. If each unit of type A brings a profit of Rs 40 and that of type B Rs 50, formulate LPP to maximize profit.
Solve the following L.P.P. by graphical method :
Maximize : Z = 7x + 11y subject to 3x + 5y ≤ 26, 5x + 3y ≤ 30, x ≥ 0, y ≥ 0.
Choose the correct alternative:
The value of objective function is maximize under linear constraints.
State whether the following is True or False :
The region represented by the inqualities x ≤ 0, y ≤ 0 lies in first quadrant.
Graphical solution set of x ≤ 0, y ≥ 0 in xy system lies in second quadrant.
Solve the following problem :
Minimize Z = 2x + 3y Subject to x – y ≤ 1, x + y ≥ 3, x ≥ 0, y ≥ 0
Solve the following problem :
A company manufactures bicyles and tricycles, each of which must be processed through two machines A and B Maximum availability of machine A and B is respectively 120 and 180 hours. Manufacturing a bicycle requires 6 hours on machine A and 3 hours on machine B. Manufacturing a tricycle requires 4 hours on machine A and 10 hours on machine B. If profits are ₹ 180 for a bicycle and ₹ 220 on a tricycle, determine the number of bicycles and tricycles that should be manufacturing in order to maximize the profit.
Solve the following problem :
A firm manufacturing two types of electrical items A and B, can make a profit of ₹ 20 per unit of A and ₹ 30 per unit of B. Both A and B make use of two essential components, a motor and a transformer. Each unit of A requires 3 motors and 2 transformers and each unit of B requires 2 motors and 4 transformers. The total supply of components per month is restricted to 210 motors and 300 transformers. How many units of A and B should be manufacture per month to maximize profit? How much is the maximum profit?
Choose the correct alternative:
If LPP has optimal solution at two point, then
Choose the correct alternative:
The corner points of feasible region for the inequations, x + y ≤ 5, x + 2y ≤ 6, x ≥ 0, y ≥ 0 are
Choose the correct alternative:
The corner points of the feasible region are (4, 2), (5, 0), (4, 1) and (6, 0), then the point of minimum Z = 3.5x + 2y = 16 is at
State whether the following statement is True or False:
If LPP has two optimal solutions, then the LPP has infinitely many solutions
State whether the following statement is True or False:
Corner point method is most suitable method for solving the LPP graphically
State whether the following statement is True or False:
The point (6, 4) does not belong to the feasible region bounded by 8x + 5y ≤ 60, 4x + 5y ≤ 40, 0 ≤ x, y
State whether the following statement is True or False:
The graphical solution set of the inequations 0 ≤ y, x ≥ 0 lies in second quadrant
The feasible region represented by the inequations x ≥ 0, y ≤ 0 lies in ______ quadrant.
If the feasible region is bounded by the inequations 2x + 3y ≤ 12, 2x + y ≤ 8, 0 ≤ x, 0 ≤ y, then point (5, 4) is a ______ of the feasible region
A company manufactures 2 types of goods P and Q that requires copper and brass. Each unit of type P requires 2 grams of brass and 1 gram of copper while one unit of type Q requires 1 gram of brass and 2 grams of copper. The company has only 90 grams of brass and 80 grams of copper. Each unit of types P and Q brings profit of ₹ 400 and ₹ 500 respectively. Find the number of units of each type the company should produce to maximize its profit
A dealer deals in two products X and Y. He has ₹ 1,00,000/- to invest and space to store 80 pieces. Product X costs ₹ 2500/- and product Y costs ₹ 1000/- per unit. He can sell the items X and Y at respective profits of ₹ 300 and ₹ 90. Construct the LPP and find the number of units of each product to be purchased to maximize its profit
A company manufactures two types of ladies dresses C and D. The raw material and labour available per day is given in the table.
| Resources | Dress C(x) | Dress D(y) | Max. availability |
| Raw material | 5 | 4 | 60 |
| Labour | 5 | 3 | 50 |
P is the profit, if P = 50x + 100y, solve this LPP to find x and y to get the maximum profit
A chemist has a compound to be made using 3 basic elements X, Y, Z so that it has at least 10 litres of X, 12 litres of Y and 20 litres of Z. He makes this compound by mixing two compounds (I) and (II). Each unit compound (I) had 4 litres of X, 3 litres of Y. Each unit compound (II) had 1 litre of X, 2 litres of Y and 4 litres of Z. The unit costs of compounds (I) and (II) are ₹ 400 and ₹ 600 respectively. Find the number of units of each compound to be produced so as to minimize the cost
A wholesale dealer deals in two kinds of mixtures A and B of nuts. Each kg of mixture A contains 60 grams of almonds, 30 grams of cashew and 30 grams of hazel nuts. Each kg of mixture B contains 30 grams of almonds, 60 grams of cashew and 180 grams of hazel nuts. A 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 and 540 grams of hazel nuts. Mixture A costs ₹ 8 and B costs ₹ 12 per kg. How many kgs of each mixture should he use to minimize the cost of the kgs
Maximize Z = 2x + 3y subject to constraints
x + 4y ≤ 8, 3x + 2y ≤ 14, x ≥ 0, y ≥ 0.
Minimize Z = 24x + 40y subject to constraints
6x + 8y ≥ 96, 7x + 12y ≥ 168, x ≥ 0, y ≥ 0
Minimize Z = x + 4y subject to constraints
x + 3y ≥ 3, 2x + y ≥ 2, x ≥ 0, y ≥ 0
Solve the LPP graphically:
Minimize Z = 4x + 5y
Subject to the constraints 5x + y ≥ 10, x + y ≥ 6, x + 4y ≥ 12, x, y ≥ 0
Solution: Convert the constraints into equations and find the intercept made by each one of it.
| Inequations | Equations | X intercept | Y intercept | Region |
| 5x + y ≥ 10 | 5x + y = 10 | ( ___, 0) | (0, 10) | Away from origin |
| x + y ≥ 6 | x + y = 6 | (6, 0) | (0, ___ ) | Away from origin |
| x + 4y ≥ 12 | x + 4y = 12 | (12, 0) | (0, 3) | Away from origin |
| x, y ≥ 0 | x = 0, y = 0 | x = 0 | y = 0 | 1st quadrant |
∵ Origin has not satisfied the inequations.
∴ Solution of the inequations is away from origin.
The feasible region is unbounded area which is satisfied by all constraints.
In the figure, ABCD represents
The set of the feasible solution where
A(12, 0), B( ___, ___ ), C ( ___, ___ ) and D(0, 10).
The coordinates of B are obtained by solving equations
x + 4y = 12 and x + y = 6
The coordinates of C are obtained by solving equations
5x + y = 10 and x + y = 6
Hence the optimum solution lies at the extreme points.
The optimal solution is in the following table:
| Point | Coordinates | Z = 4x + 5y | Values | Remark |
| A | (12, 0) | 4(12) + 5(0) | 48 | |
| B | ( ___, ___ ) | 4( ___) + 5(___ ) | ______ | ______ |
| C | ( ___, ___ ) | 4( ___) + 5(___ ) | ______ | |
| D | (0, 10) | 4(0) + 5(10) | 50 |
∴ Z is minimum at ___ ( ___, ___ ) with the value ___
A linear function z = ax + by, where a and b are constants, which has to be maximised or minimised according to a set of given condition is called a:-
If z = 200x + 500y .....(i)
Subject to the constraints:
x + 2y ≥ 10 .......(ii)
3x + 4y ≤ 24 ......(iii)
x, 0, y ≥ 0 ......(iv)
At which point minimum value of Z is attained.
