MyQuestionIcon

1

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Graphical Method of Solving Linear Programming Problems

Solve the fol...

Question

Solve the following systems of inequations graphically:

(i) $2 x + y \geq 8, x + 2 y \geq 8, x + y \leq 6$

(ii) $12 x + 12 y \leq 840, 3 x + 6 y \leq 300, 8 x + 4 y \leq 480 x \geq 0, y \geq 0$

(iii) $x + 2 y \leq 40, 3 x + y \geq 30, 4 x + 3 y \geq 60, x \geq 0, y \geq 0$

(iv) $5 x + y \geq 10, 2 x + 2 y \geq 12, x + 4 y \geq 12, x \geq 0, y \geq 0$

Open in App

Solution

(i) We have,

$2 x + y \geq 8, x + 2 y \geq 8, x + y \leq 6$

Converting the given inequation into equations,we obtain,

2x+y=8,x+2y=8,and x+y=6

Region represented by $2 x + y \geq 8$

Putting x=0 in2x+y=8,we get y=8

Putting y=0 in 2x+y=8,we get x= $\frac{8}{2} = 4$

$∴$ The line 2x+y=8 meets the coordinates axes at (0,8) and (4,0).Join these points by a thick line.

Now ,putting x=0 and y=0 in $2 x + y \geq 8$ ,we get $0 \geq 8$

This is not possible.

$∴$ We find that (0,0) is not satisifes the inequation $2 x + y \geq 8$

So,the portion not containing the origin is represented by the given inequation

Region represented by $x + 2 y \geq 8$

Putting x=0 in x+2y=8,we get y= $\frac{8}{2} = 4$

Putting y=0 in x+2y=8,we get x=8

$∴$ The line x+2y=8 meets the coordinates axes at (0,4) and (8,0).Joining these points by a thick line.

Now ,putting x=0 and y=0 in $x + 2 y \geq 8$ ,we get $0 \geq 8$

This is not possible.

$∴$ We find that (0,0) is not satisifes the inequation $x + 2 y \geq 8$

So,the portion not containing the origin is represented by the given inequation.

Region represented by $x + y \leq 6$

Putting x=0 in x+y=6,we get y=6

Putting y=0 in x+y=6,we get x=6

$∴$ The line x+y=6 meets the coordinates axes at (0,6) and (6,0).

Joining these points by a thick line.

Now ,putting x=0 and y=0 in $x + y \leq 6$ ,we get $0 \geq 6$

Therefore,(0,0) satisfies $x + y \leq 6$ ,so the portion containing the origin is represented by the given inequation.The common region of the above three regions represents the solution set of the given inequations as shown below:

(ii) We have,

$12 x + 12 y \leq 840, 3 x + 6 y \leq 300, 8 x + 4 y \leq 480 x \geq 0, y \geq 0$

Converting the given inequation into equations,we obtain,

12x+12y= 840, 3x+6y= 300,8x+4y=480, x = 0 and y=0

Region represented by $12 x + 12 y \leq 840$

Putting x=0 in 12x+12y= 840,we get y= $\frac{840}{12} = 70$

Putting y=0 in by $12 x + 12 y \leq 840$ ,we get x= $\frac{840}{12} = 70$

$∴$ The line 12x+12y= 840meets the coordinates axes at (0,70) and (70,0).Join these points by a thick line.,

Now ,putting x=0 and y=0 in $12 x + 12 y \leq 840$ ,we get $0 \leq 840$

Therefore,(0,0) satisfies the inequality $12 x + 12 y \leq 840$ ,

So,the portion not containing the origin is represented by the given inequation $12 x + 12 y \leq 840$

Region represented by $3 x + 6 y \leq 300$

Putting x=0 in $3 x + 6 y \leq 300$ ,we get y= $\frac{300}{6} = 50$

Putting y=0 in x= $\frac{300}{3} = 100$

$∴$ The line 3x+6y= 300meets the coordinates axes at (0,50) and (100,0).Joining these points by a thick line.

Now ,putting x=0 and y=0 in $3 x + 6 y \leq 300$ ,we get $0 \leq 300$

Therefore ,(0,0) satisfies the inequality $8 x + 4 y \leq 480$

Putting x=0 in 8x+4y=480,we get, y= $\frac{480}{4} = 120$

Putting y=0 in 8x+4y=480,we get, x= $\frac{480}{8} = 60$

$∴$ The line 8x+4y=480 meets the coordinates axes at (0,120) and (60,0).Join these points by a thick line.

Now ,putting x=0 and y=0 in $8 x + 4 y = 480$ ,we get $0 \leq 480$

Therefore,(0,0) satisfies the inequality $8 x + 4 y = 480$

So, the portion containing the origin is represented the solution set of the inequation $8 x + 4 y = 480$

Region represented by $x \geq 0 a n d y \geq 0$

Clearly $x \geq 0 a n d y \geq 0$ represents the first quadrant.

The common region of the above five regions represents the solution set of the given inequations as shown below:

(iii) We have,

$x + 2 y \leq 40, 3 x + y \geq 30, 4 x + 3 y \geq 60, x \geq 0, y \geq 0$

Converting the given inequation into equations,we obtain,

x+2y=40,3x+y=30,4x+3y=60,x=0 and y=0

Region represented by $x + 2 y \leq 40$

Putting x=0 in x+2y=40,we get y= $\frac{40}{2} = 20$

Putting y=0 in by x+2y=40,we get x=40

$∴$ The line x+2y=40,meets the coordinates axes at (0,20) and (40,0).Join these points by a thick line.

Now ,putting x=0 and y=0 in $x + 2 y \leq 40$ ,we get $0 \leq 40$

$t h e r e f o r e$ ,(0,0) satisfies the inequality $x + 2 y \leq 40$ ,

So,the portion not containing the origin is represented by the given inequation $x + 2 y \leq 40$

Region represented by $3 x + y \geq 30$

Putting x=0 in 3x+y=30,we get y=30

Putting y=0 in 3x+y=30,we get x= $\frac{30}{3} = 10$

$∴$ The line 3x+y=30,meets the coordinates axes at (0,30) and (10,0).Join these points by a thick line.

Now,putting x=0 and y=0 in $3 x + y \geq 30$ ,we get $0 \geq 30$

This is not possible.

Therefore ,(0,0) satisfies the inequality $3 x + y \geq 30$ .

So,the portion not containing the origin is represented by the given inequation $3 x + y \geq 30$ .

Region represented by $4 x + 3 y \geq 60$

Putting x=0 in 4x+3y= 60,we get, y= $\frac{60}{3} = 20$

Putting y=0 in 4x+3y= 60,we get, x= $\frac{60}{4} = 15$

$∴$ The line 4x+3y= 60,meets the coordinates axes at (0,20) and (15,0).Join these points by a thick line.

Now ,putting x=0 and y=0 in $4 x + 3 y \geq 60$

This is not possible.

Therefore,(0,0) satisfies the inequality $4 x + 3 y \geq 60$

So, the portion containing the origin is represented the solution set of the inequation \ $4 x + 3 y \geq 60$ .

Region represented by $x \geq 0 a n d y \geq 0$

Clearly $x \geq 0 a n d y \geq 0$ represents the first quadrant.

The common region of the above five regions represents the solution set of the given inequations as shown below:

(iv) We have,

$5 x + y \geq 10, 2 x + 2 y \geq 12, x + 4 y \geq 12, x \geq 0, y \geq 0$

Converting the given inequation into equations,we obtain,

5x+y=10,2x+2y=1,x+4y=12,x=0 and y=0

Region represented by $5 x + y \geq 10$

Putting x=0 in 5x+y=10,we get y=10

Putting y=0 in by 5x+y=10,we get x= $\frac{10}{5} = 2$

$∴$ The line 5x+y=10,meets the coordinates axes at (0,10) and (2,0).Join these points by a thick line.

Now ,putting x=0 and y=0 in $5 x + y \geq 10$ ,we get $0 \geq 10$

$t h e r e f o r e$ ,(0,0) satisfies the inequality $5 x + y \geq 10$ ,

So,the portion not containing the origin is represented by the given inequation $5 x + y \geq 10$

Region represented by $2 x + 2 y \geq 12$

Putting x=0 in 2x+2y=12,we get y= $\frac{12}{2} = 6$

Putting y=0 in 2x+2y=12,we get x= $\frac{12}{2} = 6$

$∴$ The line 2x+2y=12,meets the coordinates axes at (0,6) and (6,0).Join these points by a thick line.

Now,putting x=0 and y=0 in 2x+2y=12,we get $0 \geq 12$ ,which is not possible.

Therefore ,(0,0) satisfies the inequality 2x+2y=12.

So,the portion not containing the origin is represented by the given inequation 2x+2y=12.

Region represented by $x + 4 y \geq 12$

Putting x=0 in x+4y= 12,we get, y= $\frac{12}{4} = 3$

Putting y=0 in x+4y= 12,we get, x=12

$∴$ The line x+4y= 12,meets the coordinates axes at (0,3) and (12,0).Join these points by a thick line.

Now ,putting x=0 and y=0 in x+4y= 12,which is not possible.

Therefore,(0,0) satisfies the inequality $x + 4 y \geq 12$

So, the portion containing the origin is represented the solution set of the inequation $x + 4 y \geq 12$

Region represented by $x \geq 0 a n d y \geq 0$

Clearly $x \geq 0 a n d y \geq 0$ represents the first quadrant.

The common region of the above five regions represents the solution set of the given inequations as shown below:

Suggest Corrections

thumbs-up

1

Similar questions

Q. Solve the following systems of linear inequations graphically:

(i) 2x + 3y ≤ 6, 3x + 2y ≤ 6, x ≥ 0, y ≥ 0
(ii) 2x + 3y ≤ 6, x + 4y ≤ 4, x ≥ 0, y ≥ 0
(iii) x − y ≤ 1, x + 2y ≤ 8, 2x + y ≥ 2, x ≥ 0, y ≥ 0
(iv) x + y ≥ 1, 7x + 9y ≤ 63, x ≤ 6, y ≤ 5, x ≥ 0, y ≥ 0
(v) 2x + 3y ≤ 35, y ≥ 3, x ≥ 2, x ≥ 0, y ≥ 0

Solve the following systems of linear inequations graphically :

$(i) 2 x + 3 y \leq 6, 3 x + 2 y \leq 6, x \geq 0, y \geq 0 (i i) 2 x + 3 y \leq 6, x + 4 y \leq 4, x \geq 0, y \geq 0 (i i i) x + y \geq 1, x + 2 y \leq 8, 2 x + y \geq 2, x \geq 0, y \geq 0 (i v) x + y \geq 1, 7 x + 9 y \leq 63, y \leq 5, x \geq 0, y \geq 0 (v) 2 x + 3 y \leq 35, y \geq 3, x \geq 2, x \geq 0, y \geq 0$

Q. Solve the following linear programming problems graphically:
(i)

M a x i m u m Z = 4 x + y S u b j e c t t o x + y \leq 50 3 x + y \leq 90 x \geq 0, y \geq 0

M i n i m i z e Z = 200 x + 500 y S u b j e c t t o x + 2 y \geq 10 3 x + 4 y \leq 24 x \geq 0, y \geq 0

M a x, a n d M i n . Z = 3 x + 9 y S u b j e c t t o x + 3 y \leq 60 x + y \geq 10 x \leq y x \geq 0, y \geq 0

M i n i m i z e Z = 3 x + 2 y S u b j e c t t o x + y \geq 8 3 x + 5 y \leq 15 x \geq 0, y \geq 0

M i n i m i z e Z = x + 2 y S u b j e c t t o 2 x + y \geq 3 x + 2 y \geq 6 x \geq 0, y \geq 0

M i n . a n d M a x . Z = 5 x + 10 y S u b j e c t t o x + 2 y \leq 120 x + 2 y \geq 60 x - 2 y \geq 0 x \geq 0, y \geq 0

Q. Solve the following system of inequalities graphically

x + 2 y \leq 8

2 x + y \leq 8

x \geq 0

y \geq 0

Solve the following systems of inequalities graphically:

$x - 2 y \leq 3, 3 x + 4 y \geq 12, x \geq 0, y \geq 1$

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

Graphical Method of Solving LPP

MATHEMATICS

Watch in App

explore_more_icon

Explore more

Graphical Method of Solving Linear Programming Problems

Standard XII Mathematics

Join BYJU'S Learning Program

CrossIcon