Advertisements
Advertisements
प्रश्न
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?
Advertisements
उत्तर
Let x gift items of type A and y gift items of type B be produced by the person.
∴ Total profit Z = 75x + 125y
This is the objective function to be maximized.
The given information can be tabulated as shown below:
| Type A (x) | Type B (y) | Total time available (in hours) | |
| Cutter | 4 | 2 | 208 |
| Finisher | 2 | 4 | 152 |
∴ The constraints are 4x + 2y ≤ 208, 2x + 4y ≤ 152, x ≥ 0, y ≥ 0
∴ Given problem can be formulated as
Maximize Z = 75x + 125y
Subject to, 4x + 2y ≤ 208, 2x + 4y ≤ 152, x ≥ 0, y ≥ 0
To draw feasible region, construct table as follows:
| Inequality | 4x + 2y ≤ 208 | 2x + 4y ≤ 152 |
| Corresponding equation (of line) | 4x + 2y = 208 | 2x + 4y = 152 |
| Intersection of line with X-axis | (52, 0) | (76, 0) |
| Intersection of line with Y-axis | (0, 104) | (0, 38) |
| Region | Origin side | Origin side |
Shaded portion OABC is the feasible region,
whose vertices are O ≡ (0, 0),
A ≡ (52, 0), B and C ≡ (0, 38).
B is the point of intersection of the lines 4x + 2y = 208 i.e. 2x + y = 104 and 2x + 4y = 152
Solving the above equations, we get B ≡ (44, 16)
Here, the objective function is Z = 75x + 125y
∴ Z at O(0, 0) = 75(0) + 125(0) = 0
Z at A(52, 0) = 75(52) + 125(0) = 3900
Z at B(44, 16) = 75(44) + 125(16) = 5300
Z at C(0, 38) = 75(0) + 125(38) = 4750
∴ Z has maximum value 5300 at B(44, 16)
∴ Z is maximum, when x = 44, y = 16
Thus, a person should make 44 gift items of type A and 16 gift items of type B every month to obtain the best returns of ₹ 5300.
APPEARS IN
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.
A farmer has a 100 acre farm. He can sell the tomatoes, lettuce, or radishes he can raise. The price he can obtain is Rs 1 per kilogram for tomatoes, Rs 0.75 a head for lettuce and Rs 2 per kilogram for radishes. The average yield per acre is 2000 kgs for radishes, 3000 heads of lettuce and 1000 kilograms of radishes. Fertilizer is available at Rs 0.50 per kg and the amount required per acre is 100 kgs each for tomatoes and lettuce and 50 kilograms for radishes. Labour required for sowing, cultivating and harvesting per acre is 5 man-days for tomatoes and radishes and 6 man-days for lettuce. A total of 400 man-days of labour are available at Rs 20 per man-day. Formulate this problem as a LPP to maximize the farmer's total profit.
The corner points of the feasible region determined by the following system of linear inequalities:
2x + y ≤ 10, x + 3y ≤ 15, x, y ≥ 0 are (0, 0), (5, 0), (3, 4) and (0, 5). Let Z = px + qy, where p, q > 0. Condition on p and q so that the maximum of Z occurs at both (3, 4) and (0, 5) is
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.
Solve the following L.P.P. by graphical method:
Minimize: Z = 6x + 2y subject to x + 2y ≥ 3, x + 4y ≥ 4, 3x + y ≥ 3, x ≥ 0, y ≥ 0.
Choose the correct alternative:
The value of objective function is maximize under linear constraints.
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
Fill in the blank :
Graphical solution set of the in equations x ≥ 0, y ≥ 0 is in _______ quadrant
The region represented by the inequality y ≤ 0 lies in _______ quadrants.
Graphical solution set of x ≤ 0, y ≥ 0 in xy system lies in second quadrant.
Solve the following problem :
A company manufactures bicyles and tricycles, each of which must be processed through two machines A and B Maximum availability of machine A and B is respectively 120 and 180 hours. Manufacturing a bicycle requires 6 hours on machine A and 3 hours on machine B. Manufacturing a tricycle requires 4 hours on machine A and 10 hours on machine B. If profits are ₹ 180 for a bicycle and ₹ 220 on a tricycle, determine the number of bicycles and tricycles that should be manufacturing in order to maximize the profit.
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:
If LPP has two optimal solutions, then the LPP has infinitely many solutions
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:
Corner point method is most suitable method for solving the LPP graphically
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.
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
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
Maximize Z = 2x + 3y subject to constraints
x + 4y ≤ 8, 3x + 2y ≤ 14, x ≥ 0, y ≥ 0.
Amartya wants to invest ₹ 45,000 in Indira Vikas Patra (IVP) and in Public Provident fund (PPF). He wants to invest at least ₹ 10,000 in PPF and at least ₹ 5000 in IVP. If the rate of interest on PPF is 8% per annum and that on IVP is 7% per annum. Formulate the above problem as LPP to determine maximum yearly income.
Solution: Let x be the amount (in ₹) invested in IVP and y be the amount (in ₹) invested in PPF.
x ≥ 0, y ≥ 0
As per the given condition, x + y ______ 45000
He wants to invest at least ₹ 10,000 in PPF.
∴ y ______ 10000
Amartya wants to invest at least ₹ 5000 in IVP.
∴ x ______ 5000
Total interest (Z) = ______
The formulated LPP is
Maximize Z = ______ subject to
______
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:-
Graphical solution set of the inequations x ≥ 0 and y ≤ 0 lies in ______ quadrant.
