Advertisements
Advertisements
प्रश्न
Solve the following LPP by graphical method:
Maximize z = 11x + 8y, subject to x ≤ 4, y ≤ 6, x + y ≤ 6, x ≥ 0, y ≥ 0
उत्तर
To draw the feasible region, construct table as follows:
| Inequality | x ≤ 4 | y ≤ 6 | x + y ≤ 6 |
| Corresponding equation (of line) | x = 4 | y = 6 | x + y = 6 |
| Intersection of line with X-axis | (4, 0) | – | (6, 0) |
| Intersection of line with Y-axis | – | (0, 6) | (0, 6) |
| Region | Origin side | Origin side | Origin side |
Shaded portion OABC is the feasible region,
whose vertices are O(0, 0), A(4, 0), B and C(0, 6).
B is the point of intersection of the lines x = 4 and x + y = 6
Substituting x = 4 in x + y = 6, we get
y = 2
∴ B ≡ (4, 2)
Here, the objective function is Z = 11x + 8y
∴ Z at O(0, 0) = 11(0) + 8(0) = 0
Z at A(4, 0) = 11(4) + 8(0) = 44
Z at B(4, 2) = 11(4) + 8(2) = 44 + 16 = 60
Z at C(0, 6) = 11(0) + 8(6) = 48
∴ Z has maximum value 60 at B(4, 2)
∴ Z is maximum, when x = 4 and y = 2.
APPEARS IN
संबंधित प्रश्न
A firm manufactures 3 products A, B and C. The profits are Rs 3, Rs 2 and Rs 4 respectively. The firm has 2 machines and below is the required processing time in minutes for each machine on each product :
| Machine | Products | ||
| A | B | C | |
| M1 M2 |
4 | 3 | 5 |
| 2 | 2 | 4 | |
Machines M1 and M2 have 2000 and 2500 machine minutes respectively. The firm must manufacture 100 A's, 200 B's and 50 C's but not more than 150 A's. Set up a LPP to maximize the profit.
The corner points 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 Z = px + qy, where p, q > 0. Condition on p and q so that the maximum of Z occurs at both (3, 4) and (0, 5) is
A firm manufactures two products, each of which must be processed through two departments, 1 and 2. The hourly requirements per unit for each product in each department, the weekly capacities in each department, selling price per unit, labour cost per unit, and raw material cost per unit are summarized as follows:
| Product A | Product B | Weekly capacity | |
| Department 1 | 3 | 2 | 130 |
| Department 2 | 4 | 6 | 260 |
| Selling price per unit | ₹ 25 | ₹ 30 | |
| Labour cost per unit | ₹ 16 | ₹ 20 | |
| Raw material cost per unit | ₹ 4 | ₹ 4 |
The problem is to determine the number of units to produce each product so as to maximize total contribution to profit. Formulate this as a LPP.
Choose the correct alternative:
The value of objective function is maximize under linear constraints.
Choose the correct alternative :
The maximum value of z = 5x + 3y. subject to the constraints
Choose the correct alternative :
The point at which the maximum value of z = x + y subject to the constraints x + 2y ≤ 70, 2x + y ≤ 95, x ≥ 0, y ≥ 0 is
Fill in the blank :
Graphical solution set of the in equations x ≥ 0, y ≥ 0 is in _______ quadrant
The constraint that a factory has to employ more women (y) than men (x) is given by _______
Solve the following problem :
Maximize Z = 60x + 50y Subject to x + 2y ≤ 40, 3x + 2y ≤ 60, 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 factory produced two types of chemicals A and B The following table gives the units of ingredients P & Q (per kg) of Chemicals A and B as well as minimum requirements of P and Q and also cost per kg. of chemicals A and B.
| Ingredients per kg. /Chemical Units | A (x) |
B (y) |
Minimum requirements in |
| P | 1 | 2 | 80 |
| Q | 3 | 1 | 75 |
| Cost (in ₹) | 4 | 6 |
Find the number of units of chemicals A and B should be produced so as to minimize the cost.
Choose the correct alternative:
The corner points of feasible region for the inequations, x + y ≤ 5, x + 2y ≤ 6, x ≥ 0, y ≥ 0 are
Smita is a diet conscious house wife, wishes to ensure certain minimum intake of vitamins A, B and C for the family. The minimum daily needs of vitamins A, B, and C for the family are 30, 20, and 16 units respectively. For the supply of the minimum vitamin requirements Smita relies on 2 types of foods F1 and F2. F1 provides 7, 5 and 2 units of A, B, C vitamins per 10 grams and F2 provides 2, 4 and 8 units of A, B and C vitamins per 10 grams. F1 costs ₹ 3 and F2 costs ₹ 2 per 10 grams. How many grams of each F1 and F2 should buy every day to keep her food bill minimum
Minimize Z = 2x + 3y subject to constraints
x + y ≥ 6, 2x + y ≥ 7, x + 4y ≥ 8, x ≥ 0, y ≥ 0
Amartya wants to invest ₹ 45,000 in Indira Vikas Patra (IVP) and in Public Provident fund (PPF). He wants to invest at least ₹ 10,000 in PPF and at least ₹ 5000 in IVP. If the rate of interest on PPF is 8% per annum and that on IVP is 7% per annum. Formulate the above problem as LPP to determine maximum yearly income.
Solution: Let x be the amount (in ₹) invested in IVP and y be the amount (in ₹) invested in PPF.
x ≥ 0, y ≥ 0
As per the given condition, x + y ______ 45000
He wants to invest at least ₹ 10,000 in PPF.
∴ y ______ 10000
Amartya wants to invest at least ₹ 5000 in IVP.
∴ x ______ 5000
Total interest (Z) = ______
The formulated LPP is
Maximize Z = ______ subject to
______
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:-
Graphical solution set of the inequations x ≥ 0 and y ≤ 0 lies in ______ quadrant.
