Advertisements
Advertisements
प्रश्न
Maximize Z = 400x + 500y subject to constraints
x + 2y ≤ 80, 2x + y ≤ 90, x ≥ 0, y ≥ 0
Advertisements
उत्तर
To draw the feasible region, construct table as follows:
| Inequality | x + 2y ≤ 80 | 2x + y ≤ 90 |
| Corresponding equation (of line) | x + 2y = 80 | 2x + y = 90 |
| Intersection of line with X-axis | (80, 0) | (45, 0) |
| Intersection of line with Y-axis | (0, 40) | (0, 90) |
| Region | Origin side | Origin side |
Shaded portion OABC is the feasible region, whose vertices are O(0, 0), A(45, 0), B and C(0, 40).
B is the point of intersection of the lines 2x + y = 90 and x + 2y = 80.
Solving the above equations, we get
x = `100/3`, y = `70/3`
∴ B = `(100/3, 70/3)`
Here, the objective function is
Z = 400x + 500y
∴ Z at O(0, 0) = 400(0) + 500(0) = 0
Z at A(45, 0) = 400(45) + 500(0) = 18000
X at B`(100/3, 70/3) = 400(100/3) + 500(70/3)`
= `40000/3 + 35000/3`
= `75000/3`
= 25000
Z at C(0, 40) = 400(0) + 500(40)
= 20000
∴ Z has maximum value 25000 at x = `100/3` and y = `70/3`.
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.
A farmer has a 100 acre farm. He can sell the tomatoes, lettuce, or radishes he can raise. The price he can obtain is Rs 1 per kilogram for tomatoes, Rs 0.75 a head for lettuce and Rs 2 per kilogram for radishes. The average yield per acre is 2000 kgs for radishes, 3000 heads of lettuce and 1000 kilograms of radishes. Fertilizer is available at Rs 0.50 per kg and the amount required per acre is 100 kgs each for tomatoes and lettuce and 50 kilograms for radishes. Labour required for sowing, cultivating and harvesting per acre is 5 man-days for tomatoes and radishes and 6 man-days for lettuce. A total of 400 man-days of labour are available at Rs 20 per man-day. Formulate this problem as a LPP to maximize the farmer's total profit.
Solve the following LPP by graphical method:
Maximize z = 11x + 8y, subject to x ≤ 4, y ≤ 6, x + y ≤ 6, x ≥ 0, y ≥ 0
Solve the following L.P.P. by graphical method :
Minimize : Z = 7x + y subject to 5x + y ≥ 5, x + y ≥ 3, x ≥ 0, y ≥ 0.
Choose the correct alternative :
The maximum value of z = 5x + 3y. subject to the constraints
The region represented by the inequalities 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 :
Maximize Z = 5x1 + 6x2 Subject to 2x1 + 3x2 ≤ 18, 2x1 + x2 ≤ 12, x ≥ 0, x2 ≥ 0
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.
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.
Solve the following problem :
A person makes two types of gift items A and B requiring the services of a cutter and a finisher. Gift item A requires 4 hours of cutter's time and 2 hours of finisher's time. B requires 2 hours of cutters time, 4 hours of finishers time. The cutter and finisher have 208 hours and 152 hours available times respectively every month. The profit of one gift item of type A is ₹ 75 and on gift item B is ₹ 125. Assuming that the person can sell all the items produced, determine how many gift items of each type should be make every month to obtain the best returns?
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
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:
If the corner points of the feasible region are `(0, 7/3)`, (2, 1), (3, 0) and (0, 0), then the maximum value of Z = 4x + 5y is 12
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
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
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
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 following LPP graphically:
Maximize Z = 9x + 13y subject to constraints
2x + 3y ≤ 18, 2x + y ≤ 10, x ≥ 0, y ≥ 0
Solution: Convert the constraints into equations and find the intercept made by each one of it.
| Inequation | Equation | X intercept | Y intercept | Region |
| 2x + 3y ≤ 18 | 2x + 3y = 18 | (9, 0) | (0, ___) | Towards origin |
| 2x + y ≤ 10 | 2x + y = 10 | ( ___, 0) | (0, 10) | Towards origin |
| x ≥ 0, y ≥ 0 | x = 0, y = 0 | X axis | Y axis | ______ |
The feasible region is OAPC, where O(0, 0), A(0, 6),
P( ___, ___ ), C(5, 0)
The optimal solution is in the following table:
| Point | Coordinates | Z = 9x + 13y | Values | Remark |
| O | (0, 0) | 9(0) + 13(0) | 0 | |
| A | (0, 6) | 9(0) + 13(6) | ______ | |
| P | ( ___,___ ) | 9( ___ ) + 13( ___ ) | ______ | ______ |
| C | (5, 0) | 9(5) + 13(0) | ______ |
∴ Z is maximum at __( ___, ___ ) with the value ___.
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 ___
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.
