Advertisements
Advertisements
प्रश्न
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.
पर्याय
56
`65
55
66
उत्तर
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
संबंधित प्रश्न
Amit's mathematics teacher has given him three very long lists of problems with the instruction to submit not more than 100 of them (correctly solved) for credit. The problem in the first set are worth 5 points each, those in the second set are worth 4 points each, and those in the third set are worth 6 points each. Amit knows from experience that he requires on the average 3 minutes to solve a 5 point problem, 2 minutes to solve a 4 point problem, and 4 minutes to solve a 6 point problem. Because he has other subjects to worry about, he can not afford to devote more than
Solve the following L.P.P. by graphical method :
Maximize: Z = 3x + 5y subject to x + 4y ≤ 24, 3x + y ≤ 21, x + y ≤ 9, x ≥ 0, y ≥ 0 also find maximum value of Z.
Choose the correct alternative :
The maximum value of z = 5x + 3y. subject to the constraints
Fill in the blank :
The region represented by the in equations x ≤ 0, y ≤ 0 lines in _______ quadrants.
The constraint that a factory has to employ more women (y) than men (x) is given by _______
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 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 point at which the maximum value of Z = 4x + 6y subject to the constraints 3x + 2y ≤ 12, x + y ≥ 4, x ≥ 0, y ≥ 0 is obtained at the point
State whether the following statement is True or False:
A convex set includes the points but not the segment joining the points
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)
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
Maximize Z = 5x + 10y subject to constraints
x + 2y ≤ 10, 3x + y ≤ 12, x ≥ 0, y ≥ 0
Maximize Z = 400x + 500y subject to constraints
x + 2y ≤ 80, 2x + y ≤ 90, 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 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 ___.
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:-
