Advertisements
Advertisements
Question
Choose the correct alternative :
The maximum value of z = 10x + 6y, subjected to the constraints 3x + y ≤ 12, 2x + 5y ≤ 34, x ≥ 0, y ≥ 0 is.
Options
56
`65
55
66
Solution
Z = 10x + 6y
The given inequalities are 3x + y ≤ 12 and 2x + 5y ≤ 34.
Consider line L1 and L2 where
L1 : 3x + y = 12, L2 : 2x + 5y = 34
For line L1, plot A (0, 12) and B (4, 0)
For line L2, plot P (0, 6.8) and Q (17, 0)
Solving both lines, we get x = 2, y = 6.
The coordinates of origin O (0, 0) satisfies both the inequalities.
∴ The required region is on the origin side of both the lines L1 and L2.
As x ≥ 0, y ≥ 0, the feasible region is in the 1st quadrant.
OBRPO is the required feasible region.
At O (0, 0), Z = 0
At B (4, 0), Z = 10 (4) + 0 = 40
At R (2, 6), Z = 10 (2) + 6 (6) = 56
At P (0, 6.8), Z = 0 + 6 (6.8) = 40.8
The maximum value of Z is 56 and it occurs at R (2, 6).
APPEARS IN
RELATED QUESTIONS
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 L.P.P. by graphical method:
Maximize: Z = 10x + 25y
subject to 0 ≤ x ≤ 3,
0 ≤ y ≤ 3,
x + y ≤ 5.
Also find the maximum value of z.
Solve the following L.P.P. by graphical method:
Minimize: Z = 6x + 2y subject to x + 2y ≥ 3, x + 4y ≥ 4, 3x + y ≥ 3, x ≥ 0, y ≥ 0.
Choose the correct alternative :
The maximum value of z = 5x + 3y. subject to the constraints
Solve the following problem :
Maximize Z = 4x1 + 3x2 Subject to 3x1 + x2 ≤ 15, 3x1 + 4x2 ≤ 24, x1 ≥ 0, x2 ≥ 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 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:
The maximum value of Z = 3x + 5y subjected to the constraints x + y ≤ 2, 4x + 3y ≤ 12, x ≥ 0, y ≥ 0 is
State whether the following statement is True or False:
If the corner points of the feasible region are (0, 10), (2, 2) and (4, 0), then the minimum value of Z = 3x + 2y is at (4, 0)
The feasible region represented by the inequations x ≥ 0, y ≤ 0 lies in ______ quadrant.
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
Maximize Z = 400x + 500y subject to constraints
x + 2y ≤ 80, 2x + y ≤ 90, x ≥ 0, y ≥ 0
Minimize Z = 2x + 3y subject to constraints
x + y ≥ 6, 2x + y ≥ 7, x + 4y ≥ 8, 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 ___
Maximised value of z in z = 3x + 4y, subject to constraints : x + y ≤ 4, x ≥ 0. y ≥ 0
Shraddho wants to invest at most ₹ 25,000/- in saving certificates and fixed deposits. She wants to invest at least ₹ 10,000/- in saving certificate and at least ₹ 15,000/- in fixed deposits. The rate of interest on saving certificate is 5% and that on fixed deposits is 7% per annum. Formulate the above problem as LPP to determine maximum income yearly.
