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प्रश्न
Graphical solution set of the inequations x ≥ 0 and y ≤ 0 lies in ______ quadrant.
उत्तर
Graphical solution set of the inequations x ≥ 0 and y ≤ 0 lies in 4th quadrant.
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संबंधित प्रश्न
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.
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 :
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 region represented by the inequality y ≤ 0 lies in _______ quadrants.
The constraint that a factory has to employ more women (y) than men (x) is given by _______
Solve the following problem :
Minimize Z = 4x + 2y Subject to 3x + y ≥ 27, x + y ≥ 21, x ≥ 0, y ≥ 0
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 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 Company produces mixers and processors Profit on selling one mixer and one food processor is ₹ 2000 and ₹ 3000 respectively. Both the products are processed through three machines A, B, C The time required in hours by each product and total time available in hours per week on each machine are as follows:
| Machine/Product | Mixer per unit | Food processor per unit | Available time |
| A | 3 | 3 | 36 |
| B | 5 | 2 | 50 |
| C | 2 | 6 | 60 |
How many mixers and food processors should be produced 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?
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:
The minimum value of Z = 4x + 5y subjected to the constraints x + y ≥ 6, 5x + y ≥ 10, x, y ≥ 0 is
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:
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, 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:
Of all the points of feasible region, the optimal value is obtained at the boundary of the feasible region
A set of values of variables satisfying all the constraints of LPP is known as ______
The feasible region represented by the inequations x ≥ 0, y ≤ 0 lies in ______ quadrant.
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 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 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
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 ___
