Advertisements
Advertisements
प्रश्न
Maximize Z = 60x + 50y Subject to x + 2y ≤ 40, 3x + 2y ≤ 60, x ≥ 0, y ≥ 0
Advertisements
उत्तर
To find the graphical solution, construct the table as follows:
| Inequation | Equation | Double intercept form | Points (x, y) | Region |
| x + 2y ≤ 40 | x + 2y = 40 | `x/(40) + y/(20)` = 1 | A (40, 0) B (0, 20) |
0 + 2(0) ≤ 40 ∴ 0 ≤ 40 ∴ origin-side |
| 3x + 2y ≤ 60 | 3x + 2y = 60 | `x/(20) + y/(30)` = 1 | C (20, 0) D (0, 30) |
3(0) + 2(0) ≤ 60 ∴ 0 ≤ 60 ∴ origin-side |
| x ≥ 0 | x = 0 | – | – | R.H.S. of Y-axis |
| y ≥ 0 | y = 0 | – | – | Above X-axis |
Shaded portion OBEC is the feasible region
Whose vertices are O (0, 0), B(0, 20), E and C (20, 0)
E is the point of intersection of lines
x + 2y = 40 …(i)
3x + 2y = 60 …(ii)
∴ By (i) – (ii), we get
x + 2y = 40
3x + 2y = 60
– – –
–2x = –20
∴ x = `(-20)/(-2)`
∴ x = 10
Substituting x = 10 in (i), we get
10 + 2y = 40
∴ 2y = 40 – 10
∴ 2y = 30
∴ y = `(30)/(2)` = 15
∴ E = (10, 15)
Here, the objective function is Z = 60x + 50y
Now, we will find maximum value of Z as follows:
| Feasible points | The value of Z = 60x + 50y |
| O (0, 0) | Z = 60(0) + 50(0) = 0 |
| B (0, 20) | Z = 60(0) + 50(20) = 1000 |
| E (10, 15) | Z = 60(10) + 50(15) = 600 + 750 = 1350 |
| C (20, 0) | Z = 60(20) + 50(0) = 1200 |
∴ Z has maximum value 1350 at E (10, 15)
∴ Z is maximum, when x = 10, y = 15.
संबंधित प्रश्न
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.
Solve the following L.P.P. 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 :
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.
Solve the following L.P.P. by graphical method :
Minimize : Z = 7x + y subject to 5x + y ≥ 5, x + y ≥ 3, x ≥ 0, y ≥ 0.
Choose the correct alternative :
The point at which the maximum value of z = x + y subject to the constraints x + 2y ≤ 70, 2x + y ≤ 95, x ≥ 0, y ≥ 0 is
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 :
Maximize Z = 5x1 + 6x2 Subject to 2x1 + 3x2 ≤ 18, 2x1 + x2 ≤ 12, x ≥ 0, x2 ≥ 0
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?
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 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:
The maximum value of Z = 5x + 3y subjected to constraints 3x + y ≤ 12, 2x + 3y ≤ 18, 0 ≤ x, y is 20
State whether the following statement is True or False:
If LPP has two optimal solutions, then the LPP has infinitely many solutions
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)
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
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.
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 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 = 2x + 3y subject to constraints
x + 4y ≤ 8, 3x + 2y ≤ 14, x ≥ 0, y ≥ 0.
Maximize Z = 5x + 10y subject to constraints
x + 2y ≤ 10, 3x + y ≤ 12, 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.
Graphical solution set of the inequations x ≥ 0 and y ≤ 0 lies in ______ quadrant.
