Advertisements
Advertisements
प्रश्न
Solve each of the following inequations graphically using XY-plane:
- 11x - 55 ≤ 0
Advertisements
उत्तर
Consider the line whose equation is - 11x - 55 ≤ 0 i.e. x = -5
This represents a line parallel to Y-axis passing through the point `(-5, 0)`
Draw the line x = - 5
To find the solution set, we have to check the position of the origin (0, 0).
When x = 0, - 11x - 55 = - 11(0) - 55 = - 55 > 0
∴ the coordinates of the origin does not satisfy the given inequality.
∴ the solution set consists of the line x = -5 and the non-origin side of the line which is shaded in the graph.

APPEARS IN
संबंधित प्रश्न
Find the feasible solution of the following inequations:
x - 2y ≤ 2, x + y ≥ 3, - 2x + y ≤ 4, x ≥ 0, y ≥ 0
A company produces two types of articles A and B which requires silver and gold. Each unit of A requires 3 gm of silver and 1 gm of gold, while each unit of B requires 2 gm of silver and 2 gm of gold. The company has 6 gm of silver and 4 gm of gold. Construct the inequations and find feasible solution graphically.
A manufacturing firm produces two types of gadgets A and B, which are first processed in the foundry and then sent to the machine shop for finishing. The number of man-hours of labour required in each shop for production of A and B per unit and the number of man-hours available for the firm is as follows:
| Gadgets | Foundry | Machine shop |
| A | 10 | 5 |
| B | 6 | 4 |
| Time available (hour) | 60 | 35 |
Profit on the sale of A is ₹ 30 and B is ₹ 20 per units. Formulate the L.P.P. to have maximum profit.
A company manufactures two types of chemicals Aand B. Each chemical requires two types of raw material P and Q. The table below shows number of units of P and Q required to manufacture one unit of A and one unit of B and the total availability of P and Q.
| Chemical→ | A | B | Availability |
| Raw Material ↓ | |||
| P | 3 | 2 | 120 |
| Q | 2 | 5 | 160 |
The company gets profits of ₹ 350 and ₹ 400 by selling one unit of A and one unit of B respectively. (Assume that the entire production of A and B can be sold). How many units of the chemicals A and B should be manufactured so that the company gets a maximum profit? Formulate the problem as LPP to maximize profit.
A printing company prints two types of magazines A and B. The company earns ₹ 10 and ₹ 15 in magazines A and B per copy. These are processed on three Machines I, II, III. Magazine A requires 2 hours on Machine I, 5 hours on Machine II, and 2 hours on machine III. Magazine B requires 3 hours on machine I, 2 hours on machine II and 6 hours on Machine III. Machines I, II, III are available for 36, 50, and 60 hours per week respectively. Formulate the LPP to determine weekly production of magazines A and B, so that the total profit is maximum.
Solve the following LPP by graphical method:
Maximize z = 4x + 6y, subject to 3x + 2y ≤ 12, x + y ≥ 4, x, y ≥ 0.
Solve the following LPP by graphical method:
Maximize z = 7x + 11y, subject to 3x + 5y ≤ 26, 5x + 3y ≤ 30, x ≥ 0, y ≥ 0.
Solve the following L.P.P. by graphical method:
Minimize: z = 8x + 10y
Subject to: 2x + y ≥ 7, 2x + 3y ≥ 15, y ≥ 2, x ≥ 0, y ≥ 0.
Select the appropriate alternatives for each of the following question:
The value of objective function is maximum under linear constraints
Objective function of LPP is ______.
The point of which the maximum value of x + y subject to the constraints x + 2y ≤ 70, 2x + y ≤ 95, x, ≥ 0, y ≥ 0 is is obtained at ______.
The corner points of the feasible solution given by the inequation x + y ≤ 4, 2x + y ≤ 7, x ≥ 0, y ≥ 0 are ______.
If the corner points of the feasible solution are (0, 10), (2, 2) and (4, 0), then the point of minimum z = 3x + 2y is ______.
The half-plane represented by 3x + 2y < 8 contains the point ______.
The half-plane represented by 4x + 3y >14 contains the point ______.
Solve the following LPP:
Maximize z = 6x + 10y subject to 3x + 5y ≤ 10, 5x + 3y ≤ 15, x ≥ 0, y ≥ 0.
Solve the following LPP:
Maximize z = 2x + 3y subject to x - y ≥ 3, x ≥ 0, y ≥ 0.
Solve each of the following inequations graphically using XY-plane:
4x - 18 ≥ 0
Find graphical solution for the following system of linear in equation:
3x + 4y ≤ 12, x - 2y ≥ 2, y ≥ - 1
Solve the following LPP:
Maximize z = 4x1 + 3x2 subject to
3x1 + x2 ≤ 15, 3x1 + 4x2 ≤ 24, x1 ≥ 0, x2 ≥ 0.
Solve the following LPP:
Maximize z =60x + 50y subject to
x + 2y ≤ 40, 3x + 2y ≤ 60, x ≥ 0, y ≥ 0.
A firm manufactures two products A and B on which profit earned per unit are ₹ 3 and ₹ 4 respectively. Each product is processed on two machines M1 and M2. The product A requires one minute of processing time on M1 and two minutes of processing time on M2, B requires one minute of processing time on M1 and one minute of processing time on M2. Machine M1 is available for use for 450 minutes while M2 is available for 600 minutes during any working day. Find the number of units of products A and B to be manufactured to get the maximum profit.
A company manufactures two types of chemicals A and B. Each chemical requires two types of raw material P and Q. The table below shows number of units of P and Q required to manufacture one unit of A and one unit of B.
| Raw Material \Chemical | A | B | Availability |
| p | 3 | 2 | 120 |
| Q | 2 | 5 | 160 |
The company gets profits of ₹ 350 and ₹ 400 by selling one unit of A and one unit of B respectively. Formulate the problem as L.P.P. to maximize the profit.
A printing company prints two types of magazines A and B. The company earns ₹ 10 and ₹ 15 on magazines A and B per copy. These are processed on three machines I, II, III. Magazine A requires 2 hours on Machine I, 5 hours on Machine II and 2 hours on Machine III. Magazine B requires 3 hours on Machine I, 2 hours on Machine II and 6 hours on Machine III. Machines I, II, III are available for 36, 50, 60 hours per week respectively. Formulate the Linear programming problem to maximize the profit.
A company manufactures two types of fertilizers F1 and F2. Each type of fertilizer requires two raw materials A and B. The number of units of A and B required to manufacture one unit of fertilizer F1 and F2 and availability of the raw materials A and B per day are given in the table below:
| Raw Material\Fertilizers | F1 | F2 | Availability |
| A | 2 | 3 | 40 |
| B | 1 | 4 | 70 |
By selling one unit of F1 and one unit of F2, company gets a profit of ₹ 500 and ₹ 750 respectively. Formulate the problem as L.P.P. to maximize the profit.
Solve the following L.P.P. by graphical method:
Maximize: Z = 4x + 6y
Subject to 3x + 2y ≤ 12, x + y ≥ 4, x, y ≥ 0.
Choose the correct alternative :
Which of the following is correct?
Choose the correct alternative :
The corner points of the feasible region are (0, 0), (2, 0), `(12/7, 3/7)` and (0,1) then the point of maximum z = 7x + y
The feasible region is the set of point which satisfy.
Maximize z = 10x + 25y subject to x + y ≤ 5, 0 ≤ x ≤ 3, 0 ≤ y ≤ 3
Minimize z = 7x + y subjected to 5x + y ≥ 5, x + y ≥ 3, x ≥ 0, y ≥ 0.
Maximize z = −x + 2y subjected to constraints x + y ≥ 5, x ≥ 3, x + 2y ≥ 6, y ≥ 0 is this LPP solvable? Justify your answer.
Choose the correct alternative:
The feasible region is
Choose the correct alternative:
Z = 9x + 13y subjected to constraints 2x + 3y ≤ 18, 2x + y ≤ 10, 0 ≤ x, y was found to be maximum at the point
Constraints are always in the form of ______ or ______.
A company produces two types of pens A and B. Pen A is of superior quality and pen B is of lower quality. Profits on pens A and B are ₹ 5 and ₹ 3 per pen respectively. Raw materials required for each pen A is twice as that of pen B. The supply of raw material is sufficient only for 1000 pens per day. Pen A requires a special clip and only 400 such clips are available per day. For pen B, only 700 clips are available per day. Formulate this problem as a linear programming problem.
Solve the following linear programming problems by graphical method.
Maximize Z = 6x1 + 8x2 subject to constraints 30x1 + 20x2 ≤ 300; 5x1 + 10x2 ≤ 110; and x1, x2 ≥ 0.
Solve the following linear programming problems by graphical method.
Maximize Z = 40x1 + 50x2 subject to constraints 3x1 + x2 ≤ 9; x1 + 2x2 ≤ 8 and x1, x2 ≥ 0.
The maximum value of the objective function Z = 3x + 5y subject to the constraints x ≥ 0, y ≥ 0 and 2x + 5y ≤ 10 is:
The minimum value of the objective function Z = x + 3y subject to the constraints 2x + y ≤ 20, x + 2y ≤ 20, x > 0 and y > 0 is
Solve the following linear programming problem graphically.
Minimize Z = 200x1 + 500x2 subject to the constraints: x1 + 2x2 ≥ 10; 3x1 + 4x2 ≤ 24 and x1 ≥ 0, x2 ≥ 0.
Solve the following linear programming problem graphically.
Maximize Z = 60x1 + 15x2 subject to the constraints: x1 + x2 ≤ 50; 3x1 + x2 ≤ 90 and x1, x2 ≥ 0.
The minimum value of z = 5x + 13y subject to constraints 2x + 3y ≤ 18, x + y ≥ 10, x ≥ 0, y ≥ 2 is ______
The point which provides the solution of the linear programming problem, Max.(45x + 55y) subject to constraints x, y ≥ 0, 6x + 4y ≤ 120, 3x + 10y ≤ 180, is ______
Sketch the graph of the following inequation in XOY co-ordinate system.
2y - 5x ≥ 0
