Advertisements
Advertisements
प्रश्न
Solve the following problem :
Minimize Z = 4x + 2y Subject to 3x + y ≥ 27, x + y ≥ 21, x ≥ 0, y ≥ 0
Advertisements
उत्तर
To find the graphical solution, construct the table as follows:
| Inequation | equation | Double intercept form | Points (x1, x2) | Points (x1, x2) |
| 3x + y ≥ 27 | 3x + y = 27 | `x/(9) + y/(27)` = 1 | A (9, 0) B (0, 27) |
3(0) + 0 ≥ 27 |
| x + y ≥ 21 | x + y = 21 | `x/(21) + y/(21)` = 1 | C (21, 0) D (0, 21) |
(0) + 0 ≥ 21 ∴ 0 ≥ 21 ∴ non-origin-side |
| x ≥ 0 | x = 0 | – | R.H.S. of Y-axis | |
| y ≥ 0 | y = 0 | above X-axis |
The shaded portion CHB is the feasible region.
Whose vertices are C(21, 0), H and B(0, 27)
H is the point of intersection of lines
3x + y = 27 …(i)
x + y = 21 …(ii)
∴ By (i) – (ii), we get
3 x + y = 27
x + y = 21
– – –
2x = 6
∴ x = `(6)/(2)` = 3
Substituting x = 3 in (ii), we get
3 + y = 21
∴ y = 18
∴ H (3, 18)
Here, the objective function is Z = 4x + 2y
Now, we will find minimum value of Z as follows:
| Feasible points | The value of Z = 4x + 2y |
| C (21, 0) | Z = 4(21) + 2(0) = 84 |
| H (3, 18) | Z = 4(3) + 2(18) = 12 + 36 = 48 |
| B (0, 27) | Z = 4(0) + 2(27) = 54 |
∴ Z has minimum value 48 at H (3, 18)
∴ Z is minimum, when x = 3, y = 18
APPEARS IN
संबंधित प्रश्न
A small manufacturing firm produces two types of gadgets A and B, which are first processed in the foundry, then sent to the machine shop for finishing. The number of man-hours of labour required in each shop for the production of each unit of A and B, and the number of man-hours the firm has available per week are as follows:
| Gadget | Foundry | Machine-shop |
| A | 10 | 5 |
| B | 6 | 4 |
| Firm's capacity per week | 1000 | 600 |
The profit on the sale of A is Rs 30 per unit as compared with Rs 20 per unit of B. The problem is to determine the weekly production of gadgets A and B, so that the total profit is maximized. Formulate this problem as a LPP.
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 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.
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.
Maximize Z = 60x + 50y Subject to x + 2y ≤ 40, 3x + 2y ≤ 60, x ≥ 0, y ≥ 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 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
Choose the correct alternative:
The corner points of feasible region for the inequations, x + y ≤ 5, x + 2y ≤ 6, x ≥ 0, y ≥ 0 are
Choose the correct alternative:
The corner points of the feasible region are (4, 2), (5, 0), (4, 1) and (6, 0), then the point of minimum Z = 3.5x + 2y = 16 is at
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:
Corner point method is most suitable method for solving the LPP graphically
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
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 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 = 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
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.
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.
