Students often turn to ML Aggarwal Class 12 ISC Solutions Chapter 3 Linear Programming Chapter Test to clarify doubts and improve problem-solving skills.
ML Aggarwal Class 12 Maths Solutions Section C Chapter 3 Linear Programming Chapter Test
Question 1.
Maximize Z = 8x + 7y, subject to the constraints 3.x + y ≤ 66, x + y ≤ 45, x ≤ 20, y ≤ 40, x ≥ 0, y ≥ 0.
Answer:
To solve LPP, we convert given inequations into eqns. we draw all the lines and shade the region satisfied by these inequations.
3x + y =66 …………….(1)
x + y = 45 ………….(2)
x = 20 …(3)
y = 40 …(4)
line (1) meets coordinate axis at A (22, 0) and B (0, 66) and line (2) meets coordinate axis at C(45, 0) and D (0, 45). The line x = 20 passes through E (20, 0) and to y-axis. The line y = 40 to x-axis and pass through F (0, 40) lines (1) and (2) intersects at P\(\left(\frac{21}{2}, \frac{69}{2}\right)\); The lines (1) and (3)
intersects at Q (20, 6) and lines (1) and (4) intersects at R(\(\frac{26}{3}\), 40)
The lines y = 40 and x + y = 45 intersects at S (5, 40).
Thus, the shaded area be the bounded feasible region with corner points O (0, 0) ; E (20, 0) ; Q(20, 6); P\(\left(\frac{21}{2}, \frac{69}{2}\right)\) S (5, 40) and F (0, 40).
| Corner point | Z = 8x + 7 |
| O (0, 0) | 0 |
| E (20, 0) | 160 |
| Q (20, 6) | 160 + 42 = 202 |
| P\(\left(\frac{21}{2}, \frac{69}{2}\right)\) | 8 × \(\frac{212}{2}+\frac{69}{2}\) × 7 = 65 (Max) |
| S (5, 40) | 8 × 5 + 7 × 40 = 320 |
| F (0, 40) | 7 × 40 = 280 |
∴ Zmax = \(\frac{651}{2}\) at x = \(\frac{21}{2}\) = 10.5 and y = \(\frac{69}{2}\) = 34.5
Question 2.
A company manufactures two types of screws A and B. All the screws have to pass through a threading machine and a slotting machine. A box of type A screws requires 2 minutes on threading machine and 3 minutes on slotting machine. A box of type B screws requires 8 minutes on threading machine and 2 minutes on slotting machine. In a week, each machine is available for 60 hours. On selling these screws, the company gets a profit of? 100 per box on type A screws and ? 170 per box on type B screws. How many boxes of screws of each type should be manufactured to get maximum profit. Form it as an L.P.P. and solve graphically. (NCERT Exemplar)
Answer:
The given table can be tabulated is as follows :
| Screw/Machine | (Time) | (Time) |
| Type | Threading Machine | Slotting Machine |
| A | 2 min | 3 min |
| B | 8 min | 2 min |
| availability | 60 hours | 60 hours |
Let x boxes of screw A and y boxes of screw B are required by company to maximize his profit. Since it is given that, company gets a profit of ₹ 100 per box on type A screw’s and ₹ 170 per box on type B screws.
So profit gets by company on selling x units of boxes of screw A be 100x and profit gets by company on selling y units of boxes of screw B be 170y.
Thus, total profit of company be 100x + 170y and let Z be the total profit of company. Then Z = 100x + 170y and we want to maximise Z.
So mathematical modelling of given LPP is as under:
Max Z = 100x + 170;
Subject to constraints :
2x + 8y < 60 x 60
⇒ x + 4y < 1800
3x + 2y < 60 x 60
⇒ 3x + 2y < 3600
x, y ≥ 0
To solve LPP, we convert inequations into equations : x + 4y= 1800 ; 3x + 2y = 3600 draw all the lines and shade the region satisfied by all given inequations :
The line x + 4y = 1800 meets coordinate axis at A (1800, 0) and B (0, 450).
The line 3x + 2y = 3600 meets coordinate axis at C (1200, 0) and D (0, 1800). Both lines intersect at P (1080, 180).
Since x ≥ 0, y ≥ 0 represents the first quadrant of XOY plane.
Thus the shaded region OCPB represents the feasible and bounded region with comer points O
| Corner point | Z = 100x + 170y |
| O(0, 0) | 0 |
| C(1200, 0) | 100 × 1200 + 0= 120000 |
| P(1080, 180) | 100 × 1080 + 170 × 180 = 138,600 (Max) |
| B(0, 450) | 0 + 170 × 450 = 76,500. |
Zmax = 138,600 at x = 1080 ; y= 180
Hence required no. of boxes of type A = 1080 and required no. of boxes of type B = 180 and maximum profit = ₹ 138,600
Question 3.
A manufacturer produces two models of bikes – model X and model Y. Model X takes a 6 man-hours to make per unit, while model Y takes 10 man-hours per unit. There is a total of 450 man-hours available per week. Handling and marketing costs are ₹ 2000 and ₹ 1000 per unit for Models X and Y respectively. The total funds available for these purposes are ₹ 80000 per week. Profit per unit for models X and Y are ₹ 1000 and ₹ 600 respectively. How many bikes of each model should the manufacturer produce so as to yield a maximum profit? Form an L.P.P. and solve graphically. (NCERT Exemplar)
Answer:
The given data can be tabulated is as below :
| Bike
Model |
Man hours
(Time) |
Handling and Marketing in ₹ |
| X | 6 | 2000 |
| Y | 10 | 1000 |
| Max. availability | 450 | 80000 |
Let x bikes of model X and y bikes of model Y are required by manufacturer to maximise his profit. It is given that manufacturer gets profit per unit for model X and Y are ₹ 1000 and ₹ 600 respectively. So profit gained by manufacturer for selling* units of bike of model X andy units of bike of model Y be 1000x and 600y respectively. Let Z be the total profit of the manufacturer. Then Z = 1000x + 600y.
The mathematical modelling of given LPP is as follows :
Max Z = 1000x + 600y
Subject to constraints ; 6x + 10y ≤ 450
3x + 5y ≤ 225
and 2000x + 1000y ≤ 80000 ⇒ 2x + y ≤ 80 and x, y ≥ 0
[Since bikes of any brand cannot be negative]
To solve LPP, we convert the given in eqn’s into equations 3x + 5y = 225 and 2x + y = 80.
We draw the lines, and shaded the regions satisfied by those constraints.
The line 3x + 5y = 225 meets coordinate axis at A (75, 0) and B (0, 45).
The line 2x + y = 80 meets coordinate axis at C (40, 0) and D (0, 80) both lines intersects at P (24, 30).
The region x ≥ 0, y ≥ 0 represents the first quadrant of XOY plane.
The shaded region OCPB be the required feasible and bounded region with comer points O (0,0); C(40, 0) ; P (25, 30) and B (0, 45).
We evaluate the objective function Z at these comer points.
| Corner point | Z = 1000x + 600y |
| O (0, 0) | 0 |
| C (40, 0) | 1000 × 40 = 40,000 |
| P (25, 30) | 1000 × 25 + 600 × 30 = 43,000 (Max) |
| B (0, 45) | 1000 × 0 + 600 × 45 = 27000 |
Here, Zmax = 43,000 at x = 25 and y = 30
Hence, 25 bikes of model X and 30 bikes of model Y are sold by manufacturer to gets his maximum profit of ₹ 43,000.
Question 4.
A company sells two different products, A and B. The two products are produced in a common production process, which has a total capacity of 500 man-hours. It takes 5 hours to produce a unit of A and 3 hours to produce a unit of B. A market survey shows that maximum 70 units of A and 125 units of B can be sold. If the profit is ₹ 20 per unit for the product A and ₹ 15 per unit for the product B, how many units of each product should be sold to maximize profit ?
Answer:
Let x be the no. of units of product A and y be the no. of units of product B are required by company to maximise his profit. It is given that, company has profit of ₹ 20 per unit for product A and profit of ₹ 15 per unit for product B. Thus x units of product A and y units of product B gives profit ₹ 20x and ₹ 15y.
Let Z be the total profit of the company.
Then Z = 20x + 15y and we want to maximize Z.
Since it is given that, production processes takes 5 hours to produce product A and 3 hours to produce product B and total availability of man hours be 500 hrs. production constraint is given by 5x + 3y < 500 ; other constraints are ; x ≤ 70 ; y ≤ 125 Thus the mathematical modelling of given LPP is as follows :
Max Z = 20x + 15y
Subject to constraints ; 5x + 3y ≤ 500 ; x ≤ 70 ; y ≤ 125 and x, y ≥ 0
[since no. of products cannot be negative]
To solve LPP, we convert inequations into equations ;
5x + 3y = 500 ; x = 70 ; y = 125
We draw all the lines and shade the region satisfied by all the constraints.
The line 5x + 3y = 500 meets coordinate axis at A (100, 0) and B
The line x = 70 be a line || to y-axis and passs through C (70, 0)
The liney = 125 be a line || to x-axis and meets y-axis at D (0, 125).
The lines 5x + 3y = 500 and x = 70 meets at P (70, 50)
The lines 5x + 3y = 500 andy = 125 intersects at Q (25, 125)
The lines x = 70 andy = 125 intersects at R (70, 125)
Thus, the shaded region gives the feasible bounded region with corner points O (0, 0); C (70,0); P(70, 50); Q (25, 125) and D (0, 125).
We evaluate Z at these points
| Corner point | Z = 20x +15y |
| o (0, 0) | 0 |
| C (70, 0) | 20 × 70+ 15 × 0 = 1400 |
| P (70, 50) | 20 × 70+ 15 × 50 = 2150 |
| Q (25, 125) | 20 × 25 + 15 × 125 = 2375 |
| D (0, 125) | 20 × 0 + 15 × 125 = 1875 |
Max Z = 2375 at x = 25 and y = 125.
Thus required no. of product A = 25 and required no. of product B = 125 and Maximum profit = ₹ 2375.
Question 5.
A manufacturer of electronic circuit has a stock of 200 resistors, 120 transistors and 150 capacitors and is producing two types of circuits A and B. Type A requires 20 resistors, 10 transistors and 10 capacitors. Type B requires 10 resistors, 20 transistors and 30 capacitors. If the profit on type A circuit is ₹ 40 and that on type B circuit is ₹ 90. How many circuits of type A and of type B should be produced by the manufacturer to maximize his profit ? Formulate an L.P.P. and solve it graphically. (NCERT Exemplar)
Answer:
The given data can be tabulated as follows :
| Circuits | Resistors | Transistors | Capacitors |
| A | 20 | 10 | 10 |
| B | 10 | 20 | 30 |
| availability | 200 | 120 | 150 |
Let x be the no. of circuits of type A and y be the no. of circuits of type B produced by manufacturer to get maximum profit. It is given that, the profit on type A circuit and type B circuit be ₹ 40 and ₹ 90 respectively. So profit of manufacturer from x units of type A circuit and y units of type B circuit be ₹ 40x and ₹ 90y respectively. Let Z be the total profit of manufacturer.
Then Z = 40x + 90y, we want to maximise Z.
Hence the mathematical formulation of given LPP using table is as follows :
Max Z = 40x + 90y Subject to constraints ;
20x + 10y ≤ 200 ⇒ 2x + y ≤ 20
10x + 20y ≤ 120 ⇒ x + 2y ≤ 12
and 10x + 30y ≤ 150 ⇒ x + 3y ≤ 15
also x ≥ 0 ; y ≥ 0 [Since circuits can’t be negative]
To solve LPP, we convert given in eqn’s into equations.
2x+y = 20 …(1)
x + 2y = 12 ……(2)
and x + 3y = 15 …(3)
We draw all these lines and shaded the regions staisfied by all given inequations.
Now line (1) pass through A (10, 0) and B (0, 20). line (2) meets coordinate axis at C (12, 0) and D (0, 6). line (3) intersects coordinate axis at E (15, 0) and F (0, 5).
lines (1) and (2) intersects at P
lines (1) and (3) intersects at Q (9, 2). lines (2) and (3) intersects at R (6, 3).
The shaded region OAPRF represents the bounded feasible region with comer points O (0, 0);
A (10, 0); P\(\left(\frac{28}{3}, \frac{4}{3}\right)\) ; R (6, 3) and F (0, 5).
We evaluate Z at these comer points :
| Corner point | Z = 40x + 9y |
| o (0, 0) | 0 |
| A (10, 0) | 40 × 10 = 400 |
| P\(\left(\frac{28}{3}, \frac{4}{3}\right)\) | 40 × \(\frac{28}{3}+\frac{9 \times 4}{3}=\frac{1156}{3}\) (Max) |
| R (6, 3) | 40 × 6 + 9 × 3 = 267 |
| F (0, 5) | 40 × 0 + 9 × 5 = 45 |
Zmax = 400 at x = 10 and y = 0
Question 6.
A furniture firm manufactures chairs and tables, each requiring the use of three machines A, B and C. Production of one chair requires 2 hours on machine A, 1 hour on machine B and 1 hour on machine C. Each table requires 1 hour each on machine A and B and 3 hours on machine C. The profit obtained by selling one chair is ₹ 30 while by selling one table is ₹ 60. The total time available per week on machine A is 70 hours, on machine B is 40 hours and on machine C is 90 hours. How many chairs and tables should be made per week so as to maximize the profit ? Formulate the problem as L.P.P. and solve it graphically.
Answer:
The given data can be tablated is as follows :
| Machine
Furniture |
A
(in hr) |
B | C |
| Chair | 2 | 1 | 1 |
| Table | 1 | 1 | 3 |
| Availability | 70 | 40 | 90 |
| of time | (hr) | (hr) | (hr) |
Let x be the no. of chairs and y be no. of tables made by manufacturer to maximise his profit.
It is given that, profit gained by manufacturer in selling one chair and one table be ₹ 30 and ₹ 60 respectively.
So profit gained by manufacturer on selling x units of chairs and y units of tables be ₹ 3Ox and ₹ 60y respectively.
Let Z be the total profit of manufacturer.
Then Z = 30x + 60y and we want to maximise Z.
The mathematical formulation of given LPP using given table above is as follows :
Max Z = 30x + 60y
Subject to constraints ;
2x + y ≤ 70
x + y ≤ 40
x + 3y ≤ 0 and x, y ≥ 0 [Since tables and chairs can’t be negative] To solve LPP, we convert given inequations into equations :
2x + y = 70 …(1)
x + y = 40 …(2)
and x + 3y = 90 ……(3)
We draw all lines and shaded the regions satisfied by all inequations : eqn. (1) meets coordinate axis at A (35, 0) and B (0, 70). ‘
eqn. (2) meets coordinate axis at C (40, 0) and D (0, 40).
eqn. (3) meets coordinate axis at E (90, 0) and F (0, 30).
lines (1) and (2) intersects at P (30, 10). lines (1) and (3) intersects at Q (24, 22). lines (2) and (3) intersects at R (15, 25).
The shaded area OAPRE represents the bounded feasible region with comer points 0(0, 0); A (35, 0) ; P (30, 10) ; R (15, 25) and F (0, 30).
We evaluate Z at these corner points.
| Corner point | Z = 30x + 60y |
| O (0, 0) | 0 |
| A (35, 0) | 30 × 35 + 60 × 0 = 1050 |
| P (30, 10) | 30 × 30 + 60 × 10 = 1500 |
| R (15, 25) | 30 × 15 + 60 × 25 = 450 + 1500 |
| F (0, 30) | 30 × 0 + 60 × 30 = 1800 |
Zmax = at x = 15 and y = 25
Thus, maximum profit= ₹ 1950, when 15 chairs and 25 tables are manufactured by the manufacture.
Question 7.
Suppose every gram of wheat provides 0.1 g of proteins and 0.25 g of carbohydrates, and the corresponding values for rice are 0.05 g and 0.5 g respectively. Wheat costs ₹ 20 and rice ₹ 30 per kilogram. The minimum daily requirement of an average man for proteins and carbohydrates is 50 g and 200 g respectively. In what quantities should wheat and rice be mixed in the daily diet to provide the minimum daily requirements of proteins and carbohydrates at minimum cost ? What is the minimum cost ?
What is balanced diet and what is the importance of balanced diet in daily life? What value is being promoted ? (Value Based)
Answer:
Given data can be tabulated is as given below :
| Proteins | Carbohydrates | |
| Wheat | 01 g | 0.25 g |
| Rice | 0.05 g | 0.5 g |
| Minimum daily requirement | 50 g | 200 g |
Let x gm of wheat andy gm of rice be mixed in daily diet to provide the minimum daily requirement of proteins and carbohydrates be required to minimise the cost.
It is given that wheat costs ₹ 20 and rice costs ₹ 30 per kilogram i.e. wheat costs ₹ \(\frac{20}{1000}\) and rice 30 costs ₹ \(\frac{30}{1000}\) per gram.
So costs of x gm of wheat be ₹ \(\frac{x}{50}\) and costs of v gm of rice be ₹ \(\frac{3y}{100}\). Let Z be the total cost.
Then Z = \(\frac{x}{50}+\frac{3 y}{100}\) and we want to minimise Z.
Thus, the mathematical formulation of gien L.P.P using table is given as under :
Min Z = \(\frac{x}{50}+\frac{3 y}{100}\)
Subject to constraints ;
0.1x + 005y ≥ 50 ⇒ 2x + y ≥ 1000
and 025x + 0.5y ≥ 200 ⇒ x + 2y ≥ 800
x ≥ 0 ;y ≥ 0
[Since quantity of wheat and rice cant be negative]
To solve LPP, we convert the given inequations into eqns :
2x + y = 1000 and x + 2y = 800
We draw the given lines and shaded the region satisfied by all inequations :
The line 2x + y ≤ 1000 meets coordinate axis at A (500, 0) and B (0, 1000)
The line x + 2y = 800 meets coordinate axis at C (800, 0) and D (0, 400) both lines intersects at P (400, 200).
The shaded region OAPD represents feasible and bounded region with comer points O (0,0); A (500, 0) ; P (400, 200) and D (0, 400).
We evaluate Z at these points
| Corner point | Z = \(\frac{1200}{100}\) |
| O (0, 0) | 0 |
| A (500, 0) | Z = 10(Min) |
| P (400, 200) | Z = \(\frac{400}{50}+\frac{600}{100}\) = 14 |
| D (0, 400) | Z =\(\frac{1200}{100}\) = 12 |
Zmin = 10 at x = 500 ; y = 0 So required quantity of wheat = 500 gm and required quantity of rice = 0 gm
A balanced diet which contains all nutrients in proper quantity. Balanced diet protect us against dieseases. Values promoted are how to make life healthy.
Question 8.
To maintain one’s health, a person must fulfil certain minimum daily requirements for the following three nutrients – calcium, protein and calories. His diet consists of only food items 1 and 11 whose prices and nutrient contents are shown below :
| Food ₹ 6 per unit | Food II ₹ 10 per unit | Minimum daily requirement | |
| Calcium | 10 | 4 | 20 |
| Protein | 5 | 5 | 20 |
| Calories | 2 | 6 | 12 |
Find the combination of food items so that the cost may be minimum.
Answer:
Let x and y be the units of food I and food II respectively are used to fulfill minimum daily requirement.
The objective function which is to be minimize be Z = 6x + 10y
Subject to constraint 10x + 4y ≥ 20
i.e. 5x + 2y ≥ 10, requirement of calcium 5x + 5y ≥ 20
i.e. x + y ≥ 4, requirement of protein 2x + 6y ≥ 12
i.e. x + 3y ≥ 6, requirement of calories
For region 5x + 2y ≥ 10;
The line 5x + 2y = 10 meets coordinate axes at A(2, 0) and B(0, 5). Since (0, 0) does not satisfies 5x + 2y ≥ 10. Thus solution set of 5x + 2y ≥ 10 does not contain (0,0).
For region x + y ≥ 4; The line x + y = 4 meets coordinate axes at C(4, 0) and D(0, 4). So solution set ofx + y> 4 does not contain (0, 0) since (0, 0) does not lies on given inequality.
For region x + 3y ≥ 6; The line x + 3y = 6 meets coordinate axes at E(6, 0) and F(0, 2). So solution set of x + 3y ≥ 6 does not contain (0, 0). Since (0, 0) does not satisfies given inequality.
The lines 5x + 2v = 10 and x + y = 4 intersects at P\(\)
The lines 5x + 2y = 10 and x + 3y = 6 intersects at Q\(\)
The lines x + y = 4 and x + 3_y = 6 intersects at R(3, 1).
Also x > 0, y > 0 represents the first quadrant of region.
Thus the shaded region ERPB be the feasible region and it is unbounded.
| Corner point | Z = 6x + 10y |
| E (6, 0) | 6 × 6 + 10 × 0 = 36 |
| R(3, 1) | 6 × 3 + 10 × 1=28 (min.) |
| P\(\left(\frac{2}{3}, \frac{10}{3}\right)\) | 6 × \(\frac{2}{3}\) + 10 × \(\frac{10}{3}=\frac{112}{3}\) |
| B(0, 5) | 6 × 0 + 10 × 5 = 50 |
Here the smallest value of Z be 28 at R(3, 1). Since the feasible region is unbounded. So we have to check whether this value of Z = 28 be minimum or not. For this, we draw the line 6x + 10y = 28 and check whether the open half plane 6x + lOy < 28 have common points in the feasible region or not.
Clearly the open half plane have no common points in feasible region
∴Zmin = 28 at R(3, 1) i.e. at x = 3, y = 1.
Hence 3 units of food 1 and 1 unit of food II be required to minimise the cost.
Minimum cost = Rs. 28.
Question 9.
Two godowns A and B have a grain storage capacity of 100 quintals and 50 quintals respectively. They supply to 3 ration shops D, E and F, whose requirements are 60, 50 and 40 quintals respectively. The costs of transportation per quintal from the godowns to the shops are given in the following table:
How should the supplies be transported in order that the transportation cost is minimum? What is the minimum cost? (NCERT)
Answer:
Let godown A supply required x and y quintals of grains to the shops D and E respectively. Then, (100 – x – y) will be supplied to shop F.
It is given that the requirement at shop D is 60 quintals since x quintals are transported from godown A. Therefore, the remaining (60 – x) quintals will be transported from godown B.
Similarly, (50 -y) quintals and 40 – (100 – x – y) = (x + y – 60) quintals will be transported from godown B to shop E and F respectively.
The given problem can be represented diagrammatically as follows :
i.e. x ≥ 0, y ≥ 0 and 100 – x – y ≥ 0
i.e. x ≥ 0, y ≥ 0 and x + y ≤ 100
Also, 60 – x ≥ 0, 50 – y ≥ 0, and x + y – 60 ≥ 0
Thus, x ≤ 60, y ≤ 50, and x + y ≥ 60
Total transportation cost Z is given by,
Z = 6x + 3y + \(\frac{5}{2}\) (100 – x – y) + 4 (60 – x) + 2 (50 – y) + 3 (x + y – 60)
⇒ Z = 6x + 3y + 250 – 2.5x – 2.5y + 240 – 4x + 100 – 2y + 3x + 3y – 180
i.e. Z = 2.5x + 1.5y + 410
The mathematically modelling of given problem can be formulated as Minimize Z = 2 5x + 1.5y + 410 Subject to the constraints;
x + y ≤ 100
x ≤ 60
y ≤ 50
x + y ≥ 60
x, y ≥ 0
For region x + y ≤ 100 ; The line x + y = 100 meet coordinate axes at A(100, 0) and B(0, 100). Also (0, 0) satisfies x + y ≤ 100.; Thus the region containing (0, 0) gives the solution set of given inequation.
For region x ≤ 60 ;
The line x = 60 is a line ∥ to y-axis and meeting C(60, 0) and (0, 0) satisfies x ≤ 60. Thus, the region containing (0, 0) represents the inequation x ≤ 60.
For region y ≤ 50 ;
The line y = 50 is a line ∥ to x-axis meeting y-axis at B? (0, 50). Also (0, 0) satisfies y ≤ 50. Thus region containing (0, 0) represents the inequation y ≤ 50.
For region x + y ≥ 60 ;
The line x + y = 60 meets coordinate axes at C(60, 0) and D(0, 60). Since (0, 0) does not lies on x + y ≥ 60. Hence the region not containing (0, 0) be the solution set of x + y ≥ 30.
The line y = 50 and line x + y = 60 intersects at E(10, 50).
The line x = 60 and line x + y = 100 intersects at G (60, 40). F(50, 50) be the point of intersection of lines x = 50 and y = 50.
Thus the feasible region be the shaded area CGFE.
The corner points of feasible region are C (60, 0), G(60, 40), F(50, 50) and E(10, 50).
The values of Z at these corner points are follows :
| Corner point | Z = 2.5x + 1.5y + 410 |
| C(60, 0) | = 2.5 × 60 + 1.5 × 0 + 410 = 560 |
| G(60, 40) | = 2.5 × 60 + 1.5 × 40 + 410 = 620 |
| F(50, 50) | = 2.5 × 50 + 1.5 × 50 + 410 = 610 |
| E(10, 50) | = 2.5 × 10 + 1.50 × 50 + 410 = 510 → Minimum |
Clearly Zmin =510 and x = 10 ; y = 50
Thus, the required amount of grain transported from A to D, E, and F is 10 quintals, 50 quintals, and 40 quintals respectively and from B to D, E, and F is 50 quintals, 0 quintals, and 0 quintals respectively.
Thus, the required minimum cost is ₹ 510.