Rd Sharma XII Vol 2 2020 for Class 12 Science Maths Chapter 11 - Linear Programming

Answer:

Let x and y number of gadgets A and B respectively being produced in order to maximize the profit.

Since, each unit of gadget A takes 10 hours to be produced by machine A and 6 hours to be produced by machine B and each unit of gadget B takes 5 hours to be produced by machine A and 4 hours to be produced by machine B.

Therefore, the total time taken by the Foundry to produce x units of gadget A and y units of gadget B is $10x+6y$. This must be less than or equal to the total hours available.

Hence,  10x + 6y ≤ 1000.

This is our first constraint.

The total time taken by the machine-shop to produce x units of gadget A and y units of gadget B is 5x + 4y. This must be less than or equal to the total hours available.

Hence, 5x + 4y ≤ 600

​This is our second constraint.

Since x and y are non negative integers, therefore  $x, \mathrm{y }\ge$$0$

It is given that the profit on the sale of A is Rs 30 per unit as compared with Rs 20 per unit of B. Therefore, profit gained on x and y number of gadgets A and B is Rs 30x and Rs 20y respectively.
Let Z denotes the total cost
Therefore, Z= Rs (30x + 20y)

Hence, the above LPP can be stated mathematically as follows:

Maximize Z = 30x + 20y

subject to 

10x + 6y ≤ 1000,

5x + 4y ≤ 600

x, y ≥ 0

Answer:

Let the company produces x units of product A and y units of product B.

Since, each unit of product A costs Rs 60 and each unit of product B costs Rs 80.Therefore, x units of product A and y units of product B will cost Rs 60x and Rs 80y respectively.
Let Z denotes the total cost.

$\therefore$ Z = Rs (60x + 80y)

Also, one unit of product A requires one machine hour.

The total machine hours available with the company for product A are 400 hours.

Therefore, $x\le 400$

This is our first constraint

Also,one unit of product A and B require 1 labour hour each and there are a total of 500 labours hours.

Thus, $x+y\le 500$

​This is our second constraint.

Since, x and y are non negative integers, therefore  $x, \mathrm{y }\ge$$0$

Also, as per agreement, the company has to supply atleast 200 units of product B to its regular customers.

$\therefore y\ge 200$ 

Hence, the required LPP is  as follows:

Minimize Z = 60x + 80y

subject to 

x $\le$$400$,

x + y $\le$$500$

$y\ge 200$

$x, \mathrm{y }\ge$$0$

Answer:

Let the number of units of product A, B and C manufactured be  x, y and z respectively.

Given, machine $M_1$ takes 4 minutes to manufacture 1 unit of product A, 3 minutes to manufacture one unit of product B and 5 minute to manufacture one unit of product C.

Machine $M_2$ takes 2 minutes to manufacture 1 unit of product A, 2 minutes to manufacture one unit of product B and 4 minute to manufacture one unit of product C.

The availability is 2000 minutes for  $M_1$ and 2500 minutes for $M_2$ 

Thus, 
$$\begin{gathered} 4x+3y+5z\le 2000 \\ 2x+2y+4z\le 2500 \end{gathered}$$

Number of units of products cannot be negative.
So, $x, y, z\ge 0$

Further, it is given that the firm should manufacture 100 A's, 200 B's and 50 C's but not more than 150 A's.

Then,
$$\begin{gathered} 100\le x\le 150 \\ y\ge 200 \\ z\ge 50 \end{gathered}$$

 Let Z denotes the profit

 $\therefore \mathrm{Z}=$3x + 2y + 4z

​Hence, the required LPP is as follows :

Maximize  $\mathrm{Z}=$3x + 2y + 4z

subject to 

$$\begin{gathered} 4x+3y+5z\le 2000 \\ 2x+2y+4z\le 2500 \end{gathered}$$

$$\begin{gathered} 100\le x\le 150 \\ y\ge 200 \\ z\ge 50 \\ x, y, z\ge 0 \end{gathered}$$

 

Answer:

Let the firm produces x units of product A and y units of product B.
Since, each unit of product A requires one minute on machine $M_{\mathit{1}}$ and two minutes on machine $M_2$.
Therefore, x units of product A will require product $x$ minutes on machine ​$M_{\mathit{1}}$ and $2x$ minutes on machine $M_2$
Also, 
Since each unit of product B requires one minute on machine $M_{\mathit{1}}$ and one minute on machine $M_2$.
Therefore, $y$ units of product A will require product $y$ minutes on machine ​ $M_{\mathit{1}}$ and $y$ minutes on machine $M_2$

It is given that the machine $M_1$ is available for $6 \mathrm{hours} \mathrm{and} 40 \mathrm{minutes}$ i.e. $400 \mathrm{minutes}$ and machine $M_2$ is available for 10 hours i.e. 600 minutes

Thus, 
$$\begin{gathered} x+y\le 400 \\ 2x+y\le 600 \end{gathered}$$

Since,units of the products cannot be negative,so $x, \mathrm{y }\ge$$0$
Let Z denotes the total profit
$\mathit{\therefore }\mathit{ }\mathrm{Z}=2x+3y$ which is to be maximised

Hence, the required LPP is as follows:

Maximize Z = 2x + 3y

subject to 

$$\begin{gathered} x+y\le 400 \\ 2x+y\le 600 \end{gathered}$$

$x, \mathrm{y }\ge$$0$

Answer:

Let plant I be run for x days and plant II be run for y days

Then,
 

Answer:

Let x units of product A and y units of product B be produced.
Then,
Since, it takes 5 hours to produce a unit of A and 3 hours to produce a unit of B.
Therefore, it will take 5x hours to produce x units of A and 3y hours to produce y units of B.
As, the total capacity is of 45000 man hours.
$\Rightarrow 5x+3y\le 45000$
Also,
The maximum number of units of A that can be sold is 7000 and that of B is 10,000 and number of units cannot be negative.
Thus, $0\le x\le 7000, 0\le y\le 10000$
Now,
Total profit = $60x+40y$
Here, we need to maximize profit

Thus, the objective function will be maximize $\mathrm{Z}=60x + 40y$

​Hence, the required LPP is as follows:

Maximize Z = 60x + 40y

subject to  

$$\begin{gathered} 5x+3y\le 45000 \\ x\le 7000 \\ y\le 10000 \\ x, y\ge 0 \end{gathered}$$

Answer:

Let the person takes x lbs and y lbs of food I and II respectively that were taken in the diet.

Since, per lb of food I  costs Rs 60 and that of food II costs Rs 100.
Therefore, x lbs of food I  costs Rs 60x and y lbs of food II costs Rs 100y.

Total cost per day = Rs (60x + 100y)

​Let Z denote the total cost per day

Then, Z = 60x + 100y

Total amount of calcium in the diet is $10x+5y$

Since, each lb of food I contains 10 units of calcium.Therefore, x lbs of food I contains 10x units of calcium.
Each lb of food II contains 5 units of calciu.So,y lbs of food II contains 5y units of calcium.

Thus, x lbs of food I and y lbs of food II contains 10x + 5y units of calcium.

But, the minimum requirement is 20 lbs of calcium.

$\therefore$ $10x+5y\ge 20$

Since, each lb of food I contains 5 units of protein.Therefore, x lbs of food I contains 5x units of protein.
Each lb of food II contains 4 units of protein.So,y lbs of food II contains 4y units of protein.

Thus, x lbs of food I and y lbs of food II contains 5x + 4y units of protein.

But, the minimum requirement is 20 lbs of protein.
$\therefore$ $5x+4y \ge 20$

Since, each lb of food I contains 2 units of calories.Therefore, x lbs of food I contains 2x units of calories.
Each lb of food II contains  units of calories.So,y lbs of food II contains 6y units of calories.

Thus, x lbs of food I and y lbs of food II contains 2x + 6y units of calories.

But, the minimum requirement is 13 lbs of calories.
$\therefore 2x+6y\ge 13$

Finally, the quantities of food I and food II are non negative values.

So, $x, y \ge 0$

​Hence, the required LPP is as follows:

Min Z = 60x + 100y

subject to  

$$\begin{gathered} 10x+5y\ge 20 \\ 5x+4y\ge 20 \\ 2x+6y\ge 13 \\ x, y\ge 0 \end{gathered}$$

Answer:

Let x and y units of products A and B were manufactured respectively.
The contribution to profit is Rs 2 for each unit of A and Rs 3 for each unit of B.
Therefore for x units of A and y units of B,the contribution to profit would be Rs 2x and Rs 3y respectively.

​Let Z denote the total profit

Then, Z = Rs (2x + 3y)

Total hours required for grinding, turning, assembling and testing are  $x+2y, 3x+y, 6x+3y, 5x+4y$ respectively.

The available capacities of these operations in hours for the given period are grinding 30, turning 60, assembling 200 and testing 200.

$\therefore$ $x+2y\le 30, 3x+y\le 60, 6x+3y\le 200, 5x+4y\le 200$
Units of products cannot be negative.Therefore,
$x,y\ge 0$

Hence, the required LPP is as follows:

Maximize Z = 2x + 3y

subject to

$$\begin{gathered} x+2y\le 30, \\ 3x+y\le 60, \\ 6x+3y\le 200, \\ 5x+4y\le 200 \end{gathered}$$

Answer:

Let x and y units of food F₁ and food F₂ were mixed.
Clearly, x ≥ 0 and  y ≥ 0

One unit of food F₁ contains 2 units of vitamin A and one unit of of food F₂ contains 4 units of vitamin A. Therefore, x and y units of food F₁ and food F₂ respectively contains 2x and 4y units of vitamin A.

It is given that the minimum daily requirements for a person of vitamin A is 40 units.
Hence, 2x+ 4y ≥ 40

One unit of food F₁ contains 3 units of vitamin B and one unit of food F₂ contains 2 units of  of vitamin B. Therefore, x and y units of F₁ and F₂ respectively contains 3x and 2y units of vitamin B.

It is given that the minimum daily requirements for a person of vitamin B is 50 units.
Hence, 3x+ 2y ≥ 50

One unit of food F₁ and food F₂ cost Rs 50 and 25 respectively. Therefore, x and y units of food F₁ and food F₂ costs Rs 50x and Rs 25y respectively.
​Let Z denote the total cost

Then, Z = Rs (50x + 25y)

​Hence, the required LPP is 

Minimize Z = 50x + 25y

subject to  

2x+ 4y ≥ 40
3x+ 2y ≥ 50
x ≥ 0,y ≥ 0

Answer:

Let x number of trucks and y number of automobiles were produced to maximize the profit.
Since, the manufacturer makes profit of Rs 30000 on each truck and Rs 2000 on each automobile.
Therefore, on x number of trucks and y number of automobiles profit would be Rs 30000x and Rs 2000y respectively.
Total profit  = Rs (30000x + 2000y)

​Let Z denote the total profit
Then, Z = 30000x + 2000y

​Since, 5 man-days and 2 man-days were required to produce each truck and automobile at shop A.
Therefore, 5x man-days and 2y man-days are required to produce x trucks  and y automobiles at shop A.

Also,
 Since 3 man-days were required to produce each truck and automobile at shop B.
 Therefore, 3x man-days and 3y man-days are required to produce x trucks and y automobiles​.

As, shop A has 180 man-days per week available while shop B has 135 man-days per week.
​$\therefore$ $5x+2y\le 180, 3x+3y\le 135$

Number of trucks and automobiles cannot be negative.
$\therefore$ $x, y\ge 0$

Hence, the required LPP is as follows :

Maximize Z = 30000x + 2000y

subject to  

$$\begin{gathered} 5x+2y\le 180, \\ 3x+3y\le 135, \\ x\ge 0, y\ge 0 \end{gathered}$$

Answer:

Let x and y units of product A and B were manufactured respectively.
Labour cost per unit to manufacture product A and product B is Rs 16 and Rs 20 respectively.Therefore, labour cost for x and y units of product A and product B is Rs 16x and Rs 20y respectively.

Total labour cost to manufacture product A and product B is Rs (16x+20y)

Raw material cost per unit to manufacture product A and product B is Rs 4 and Rs 4 respectively.Therefore,raw material cost for x and y units of product A and product B is Rs 4x and Rs 4y respectively.

Total raw material cost to manufacture product A and product B is Rs (4x + 4y)
Hence, total cost price to manufacture product A and product B = Total labour cost + Total raw material cost
                                                                                                        = 16x + 4x + 20y + 4y
                                                                                                        = 20x + 24y



Selling price per unit for product A and product B is Rs 25 and Rs 30 respectively. Therefore, total selling price for product A and product B is Rs 25x and Rs 30y respectively.
Total selling price = 25x + 30y 

∴  Total profit  = Total selling price − Total cost price = 25x + 30y $-$(20x + 24y) 
                                                                                        = 5x + 6y

​Let Z denote the total profit

Then, Z = $5x+6y$ 

One unit of product A and product B requires 3 hours and 2 hours respectively at department 1.Therefore, x units and y units of product A and product B
require 3x hours and 2y hours respectively.
The weekly capacity of department 1 is 130.

$\therefore 3x+2y\le 130$

One unit of product A and B requires 4 hours and 6 hours respectively at department 2.Therefore, x units and y units of product A and product B require 4x hours and 6y hours respectively.
The weekly capacity of department  2 is 260.

$\therefore 4x+6y\le 260$

Units of products cannot be negative.Therefore,
$x, y\ge 0$


​Hence, the required LPP is as follows:

Maximize Z = 5x + 6y

subject to  

$$\begin{gathered} 3x+2y\le 130, \\ 4x+6y\le 260, \\ x\ge 0, y\ge 0 \end{gathered}$$

Answer:

Let $x$ number of model 314 planes and y  number of model 535 planes were used.

It is given that cost of one model 314 plane is Rs 100000 and cost of one model 535 plane is Rs 150000.
Therefore, cost of x model 314 plane is Rs 100000x and cost of y model 535 plane is Rs 150000y.

Total cost price = 100000x + 150000y

​​Let Z denote the total cost

Then, Z = $100000x+150000y$ 

Also, 
Each model 314 planes have 20 first class and 30 tourist class seats and each model 535 planes has 20 first class and 60 tourist class seats.

The group needs 160 first class seats and 300 tourist class seats.

$$\begin{gathered} \therefore 20x+20y\ge 160 \\ 30x+60y\ge 300 \end{gathered}$$

Number of planes cannot be negative.
Therefore, $x, y\ge 0$

Answer:

Let Amit correctly solves x problems from the first set, y problems from the second set and z problems from the third set.
Given,
Amit cannot submit more than 100 correctly solved problems.
$\therefore x+y+z\le 100$

The problem in the first set are worth 5 points each,those in the second set worth 4 points each and those in the third set worth 6 points each.
Therefore, x problems from the first set worth 5x points, y problems from the second set worth 4y points and z problems from the third set worth 6z points.
Thus, total credit points will be $5x+4y+6z$.
 
Let Z denotes the total credit of Amit

$\therefore \mathrm{Z} = 5x+4y+6z$

It requires 3 minutes to solve a 5 point problem, 2 minutes to solve a 4 point problem and 4 minutes to solve a 6 point problem.Therefore,x problems from the first set require 3x minutes, y problems from the second set require 2y minutes and z problems from the third set require 4z minutes.
Thus, the total time require by Amit will be (3x + 2y + 4z) minutes.

It is given that the total time that Amit can devote on his mathematics assignment is $3\frac{1}{2}\mathrm{hours} \mathrm{i}.\mathrm{e}. 210 \mathrm{minutes}.$

$\therefore 3x+2y+4z\le 210$

Further, it is given that the total time that Amit can devote in solving first two types of problems cannot be more than $2\frac{1}{2}$  hours​  i.e. $150 \mathrm{minutes}$.
$\therefore 3x+2y\le 150$

Number of problems cannot be negative.Therefore,

 $x, y\ge 0$

Answer:

Let the farmer sow tomatoes in x acres, lettuce in y acres & radishes in z acres of the farm.
Average yield per acre is 2000 kgs for tomatoes, 3000 kgs of lettuce and 1000 kg of radishes.

Thus, the farmer raised 2000x kg of tomatoes, 3000y kg of lettuce and 1000z kg of radishes.

Given, price he can obtain is Re 1 per kilogram for tomatoes, Re 0.75 a head for lettuce and Rs 2 per kilogram for radishes.


∴  Selling price = Rs $\left[2000x\left(1\right)+3000y\left(0.75\right)+1000z\left(2\right)\right]$ = Rs (2000x + 2250y + 2000z)

Labour required for sowing, cultvating and harvesting per acre is 5 man-days for tomatoes and radishes and 6 man-days for lettuce.Therefore, labour required for sowing, cultivating and harvesting per acre is 5x for tomatoes, 6y for lettuce and 5z for radishes. 

Number of man-days required in sowing, cultivating and harvesting= $5x+6y+5z$

Price of one man-day = Rs 20

$\therefore$ Labour cost =  $20\left(5x+6y+5z\right)=100x+120y+100z$ 

Also, fertilizer is available at Re 0.50 per kg and the  amount required per acre is 100 kgs each for tomatoes and lettuce and 50 kgs for radishes.
Therefore, fertilizer required is 100x kgs for the tomatoes sown in x acres, 100y kgs for the lettuce sown in y acres and 50z kgs for radishes sown in z acres of land. 
Hence, total fertilizer used= (100x + 100y +50z) kgs

Thus, fertilizer's cost = $\mathrm{Rs} 0.5\times \left(100x+100y+50z\right)=\mathrm{Rs}\left(50x+50y+25z\right)$

So, the total price that has been cost to farmer = Labour cost + Fertilizer cost
                                                                             = Rs $\left(150x+170y+125z\right)$
                                                                            
Profit made by farmer =  Selling price $-$ Cost price
                                     = Rs (2000x + 2250y + 2000z)​− Rs (150x + 170y + 125z)
                                     =  Rs $\left(1850x+2080y+1875z\right)$  

Let Z denotes the total profit

$\therefore \mathrm{Z} =1850x+2080y+1875z$

Now, 
Total area of the farm = 100 acres
$x+y+z\le 100$

Also, it is given that the total man-days available are  400.
Thus,  $5x+6y+5z\le 400$
Area of the land cannot be negative.
Therefore, $x, y\ge 0$

Answer:

Let the number of large vans and small vans used for transporting the packages be x and y, respectively.

It is given that the cost of engaging each large van is ₹400 and each small van is ₹200.

Cost of engaging x large vans = ₹400x

Cost of engaging y small vans = ₹200y

Let Z be the total cost of engaging x large vans and y small vans.

∴ Z = ₹(400x + 200y)

The firm has to transport at least 1200 packages daily using large vans which carry 200 packages each and small vans which can take 80 packages each.

∴ Number of packages transported by x large vehicles + Number of packages transported by y small vehicles ≥ 1200

⇒ 200x + 80y ≥ 1200

Not more than ₹3000 is to be spent daily on the transportation.

∴ 400x + 200y ≤ 3000

Also, the number of large vans cannot exceed the number of small vans.

∴ x ≤ y

Thus, the linear programming problem of the given problem is 

Minimise Z = ₹(400x + 200y)

Subject to constraints

200x + 80y ≥ 1200

400x + 200y ≤ 3000

x ≤ y

x ≥ 0, y ≥ 0

Answer:

The given data can be arranged in the following tabular form:

Answer:

The feasible region determined by the system of constraints, 3x + 5y ≤ 15, 5x + 2y ≤ 10, x ≥ 0, and y ≥ 0, are as follows.

The corner points of the feasible region are O(0, 0), C(2, 0),$E\left(\frac{20}{19},\frac{45}{19}\right)$ and B(0, 3).

The values of Z at these corner points are as follows.

Therefore, the maximum value of Z is $\frac{235}{19}\mathrm{at} \mathrm{the} \mathrm{point} \left(\frac{20}{19},\frac{45}{19}\right)$.Hence, x= $\frac{20}{19}$ and y =$\frac{45}{19}$ is the optimal solution of the given LPP.
Thus, the optimal value of Z is $\frac{235}{19}$.

Answer:

First, we will convert the given inequations into equations, we obtain the following equations:
2x + 3y = 13, 3x +y = 5, x = 0 and y = 0

Region represented by 2x + 3y ≤ 13 :
The line 2x + 3y = 13 meets the coordinate axes at $A\left(\frac{13}{2},0\right)$ and $B\left(0,\frac{13}{3}\right)$ respectively. By joining these points we obtain the line 2x + 3y = 13.
Clearly (0,0) satisfies the inequation 2x + 3y ≤ 13. So,the region containing the origin represents the solution set of the inequation 2x + 3y ≤ 13.

Region represented by 3x + y ≤ 5:
The line 5x + 2y = 10 meets the coordinate axes at $C\left(\frac{5}{3},0\right)$ and D(0, 5) respectively. By joining these points we obtain the line 3x + y = 5.
Clearly (0,0) satisfies the inequation 3x + y ≤ 5. So,the region containing the origin represents the solution set of the inequation 3x + y ≤ 5.

Region represented by x ≥ 0 and y ≥ 0:
Since, every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by the inequations x ≥ 0, and y ≥ 0.
 

The feasible region determined by the system of constraints, 2x + 3y ≤ 13, 3x + y ≤ 5, x ≥ 0, and y ≥ 0, are as follows.

The corner points of the feasible region are O(0, 0), $C\left(\frac{5}{3},0\right)$ ,$E\left(\frac{2}{7},\frac{29}{7}\right)$ and $B\left(0,\frac{13}{3}\right)$.

The values of Z at these corner points are as follows.

We see that the maximum value of the objective function Z is 15 which is at C $\left(\frac{5}{3},0\right)$ and $E\left(\frac{2}{7},\frac{29}{7}\right)$.
Thus, the optimal value of Z is 15.

 

Answer:

First, we will convert the given inequations into equations, we obtain the following equations:
4x + y = 20, 2x +3y = 30, x = 0 and y = 0

Region represented by 4x + y ≥ 20 :
The line 4x + y = 20  meets the coordinate axes at A(5, 0) and B(0, 20) respectively. By joining these points we obtain the line 4x + y = 20.
Clearly (0,0) does not satisfies the inequation 4x + y ≥ 20. So,the region in xy plane which does not contain the origin represents the solution set of the inequation 4x + y ≥ 20.

Region represented by 2x +3y ≥ 30:
The line 2x +3y = 30 meets the coordinate axes at C(15,0) and D(0, 10) respectively. By joining these points we obtain the line
2x +3y = 30.Clearly (0,0) does not satisfies the inequation 2x +3y ≥ 30. So,the region which does not contain the origin represents the solution set of the inequation 2x +3y ≥ 30.

Region represented by x ≥ 0 and y ≥ 0:
Since, every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by the inequations x ≥ 0, and y ≥ 0.
The feasible region determined by the system of constraints, 4x + y ≥ 20, 2x +3y ≥ 30, x ≥ 0, and y ≥ 0, are as follows.

The corner points of the feasible region are B(0, 20), C(15,0), E(3,8) and C(15,0).

The values of Z at these corner points are as follows.

Therefore, the minimum value of Z is 134 at the point E(3,8). Hence, x = 3 and y =8 is the optimal solution of the given LPP.
Thus, the optimal value of Z is 134.

Answer:

First, we will convert the given inequations into equations, we obtain the following equations:
2x + y = 18, 3x + 2y = 34

Region represented by 2x + y ≥ 18:
The line 2x + y = 18 meets the coordinate axes at A(9, 0) and B(0, 18) respectively. By joining these points we obtain the line 2x + y = 18.
Clearly (0,0) does not satisfies the inequation 2x + y ≥ 18. So,the region in xy plane which does not contain the origin represents the solution set of the inequation 2x + y ≥ 18.

Region represented by 3x + 2y ≤ 34:
The line 3x + 2y = 34 meets the coordinate axes at $C\left(\frac{34}{3}, 0\right)$ and D(0, 17) respectively. By joining these points we obtain the line 3x + 2y = 34.
Clearly (0,0) satisfies the inequation 3x + 2y ≤ 34. So,the region containing the origin represents the solution set of the inequation 3x + 2y ≤ 34.

The corner points of the feasible region are A(9, 0), $C\left(\frac{34}{3}, 0\right)$ and E(2, 14).

The values of Z at these corner points are as follows.

Therefore, the maximum value of Z is $\frac{1700}{3}\mathrm{at} \mathrm{the} \mathrm{point} \left(\frac{34}{3}, 0\right)$.Hence, x = $\frac{34}{3}$ and y =0 is the optimal solution of the given LPP.
Thus, the optimal value of Z is $\frac{1700}{3}$.

Answer:

We need to maximize Z = 4x + 3y

First, we will convert the given inequations into equations, we obtain the following equations:
3x + 4y = 24, 8x + 6y = 48, x = 5, y = 6, x = 0 and y = 0.

The line 3x + 4y = 24 meets the coordinate axis at A(8, 0) and B(0,6). Join these points to obtain the line 3x + 4y = 24.
Clearly, (0, 0) satisfies the inequation 3x + 4y ≤ 24.So, the region in xy-plane that contains the origin represents the solution set of the given equation.

The line 8x + 6y = 48 meets the coordinate axis at C(6, 0) and D(0,8). Join these points to obtain the line 8x + 6y = 48.
Clearly, (0, 0) satisfies the inequation 8x + 6y  ≤ 48. So, the region in xy-plane that contains the origin represents the solution set of the given equation.

x = 5 is the line passing through x = 5 parallel to the Y axis.
y = 6 is the line passing through y = 6 parallel to the X axis.

Region represented by x ≥ 0 and y ≥ 0:
Since, every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by the inequations.
These lines are drawn using a suitable scale.

The corner points of the feasible region are O(0, 0), $G\left(5, 0\right)$, $F\left(5,\frac{4}{3}\right)$, $E\left(\frac{24}{7},\frac{24}{7}\right)$ and $B\left(0, 6\right)$.

The values of Z at these corner points are as follows.
 

We see that the maximum value of the objective function Z is 24 which is at $F\left(5,\frac{4}{3}\right)$ and $E\left(\frac{24}{7},\frac{24}{7}\right)$.
Thus, the optimal value of Z is 24.

Answer:

First, we will convert the given inequations into equations, we obtain the following equations:
3x + 2y = 80, 2x + 3y = 70, x = 0 and y = 0

Region represented by 3x + 2y ≤ 80 :
The line 3x + 2y = 80 meets the coordinate axes at $A\left(\frac{80}{3}, 0\right)$ and $B\left(0, 40\right)$ respectively. By joining these points we obtain the line 3x + 2y = 80.
Clearly (0,0) satisfies the inequation 3x + 2y ≤ 80 . So,the region containing the origin represents the solution set of the inequation 3x + 2y ≤ 80 .

Region represented by 2x + 3y ≤ 70:
The line 2x + 3y = 70 meets the coordinate axes at $C\left(35, 0\right)$ and $D\left(0, \frac{70}{3}\right)$ respectively. By joining these points we obtain the line 2x + 3y ≤ 70.
Clearly (0,0) satisfies the inequation 2x + 3y ≤ 70. So,the region containing the origin represents the solution set of the inequation 2x + 3y ≤ 70.

Region represented by x ≥ 0 and y ≥ 0:
Since, every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by the inequations x ≥ 0 and y ≥ 0.
 

The feasible region determined by the system of constraints, 3x + 2y ≤ 80, 2x + 3y ≤ 70, x ≥ 0, and y ≥ 0 are as follows.

The corner points of the feasible region are O(0, 0), $A\left(\frac{80}{3}, 0\right)$ ,$E\left(20, 10\right)$ and $D\left(0,\frac{70}{3}\right)$.

The values of Z at these corner points are as follows.

We see that the maximum value of the objective function Z is 400 which is at $A\left(\frac{80}{3}, 0\right)$ and $E\left(20, 10\right)$.
Thus, the optimal value of Z is 400.

 

Answer:

First, we will convert the given inequations into equations, we obtain the following equations:
3x + y = 12, 2x + 5y = 34, x = 0 and y = 0

Region represented by 3x + y ≤ 12:
The line 3x + y = 12 meets the coordinate axes at $A\left(4, 0\right)$ and $B\left(0, 12\right)$ respectively. By joining these points we obtain the line 3x + y = 12.
Clearly (0,0) satisfies the inequation 3x + y ≤ 12. So,the region containing the origin represents the solution set of the inequation 3x + y ≤ 12 .

Region represented by 2x + 5y  ≤ 34:
The line 2x + 5y = 34 meets the coordinate axes at $C\left(17, 0\right)$ and $D\left(0, \frac{34}{5}\right)$ respectively. By joining these points we obtain the line 2x + 5y  ≤ 34.
Clearly (0,0) satisfies the inequation 2x + 5y  ≤ 34. So,the region containing the origin represents the solution set of the inequation 2x + 5y  ≤ 34.

Region represented by x ≥ 0 and y ≥ 0:
Since, every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by the inequations x ≥ 0 and y ≥ 0.
 

The feasible region determined by the system of constraints, 3x + y ≤ 12, 2x + 5y  ≤ 34, x ≥ 0, and y ≥ 0 are as follows.

The corner points of the feasible region are O(0, 0), $A\left(4, 0\right)$ ,$E\left(2, 6\right)$ and $D\left(0,\frac{34}{5}\right)$.

The values of Z at these corner points are as follows:

We see that the maximum value of the objective function Z is 56 which is at $E\left(2, 6\right)$ that means at x = 2 and y = 6.
Thus, the optimal value of Z is 56.
 

Answer:

We have to maximize Z = 3x + 4y
First, we will convert the given inequations into equations, we obtain the following equations:
2x + 2y = 80, 2x + 4y = 120

Region represented by 2x + 2y ≤ 80:
The line 2x + 2y = 80 meets the coordinate axes at $A\left(40, 0\right)$ and $B\left(0, 40\right)$ respectively. By joining these points we obtain the line 2x + 2y = 80.
Clearly (0,0) satisfies the inequation 2x + 2y ≤ 80. So,the region containing the origin represents the solution set of the inequation 2x + 2y ≤ 80.

Region represented by 2x + 4y ≤ 120:
The line 2x + 4y = 120 meets the coordinate axes at $C\left(60, 0\right)$ and $D\left(0, 30\right)$ respectively. By joining these points we obtain the line 2x + 4y ≤ 120.
Clearly (0,0) satisfies the inequation 2x + 4y ≤ 120. So,the region containing the origin represents the solution set of the inequation 2x + 4y ≤ 120.

The feasible region determined by the system of constraints, 2x + 2y ≤ 80, 2x + 4y ≤ 120 are as follows:

The corner points of the feasible region are O(0, 0), $A\left(40, 0\right)$, $E\left(20, 20\right)$ and $D\left(0, 30\right)$.

The values of Z at these corner points are as follows:

We see that the maximum value of the objective function Z is 140 which is at $E\left(20, 20\right)$ that means at x = 20 and y = 20.
Thus, the optimal value of Z is 140.
 

Answer:

We have to maximize Z = 7x + 10y
First, we will convert the given inequations into equations, we obtain the following equations:
x + y = 30000,y = 12000, x = 6000, x = y, x = 0 and y = 0.

Region represented by x + y ≤ 30000:
The line x + y = 30000 meets the coordinate axes at $A\left(30000, 0\right)$ and $B\left(0, 30000\right)$ respectively. By joining these points we obtain the line x + y = 30000.
Clearly (0,0) satisfies the inequation x + y ≤ 30000. So,the region containing the origin represents the solution set of the inequation x + y ≤ 30000.

The line y = 12000 is the line that passes through C(0,12000) and parallel to x axis.

The line x = 6000 is the line that passes through (6000, 0) and parallel to y axis.

Region represented by x ≥ y
The line x = y is the line that passes through origin.The points to the right of the line x = y satisfy the inequation x ≥ y.
Like by taking the point (−12000, 6000).Here, 6000 > −12000 which implies y > x. Hence, the points to the left of the line x = y will not satisfy the given inequation x ≥ y.

Region represented by x ≥ 0 and y ≥ 0:
Since, every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by the inequations x ≥ 0 and y ≥ 0.

The feasible region determined by the system of constraints, x + y ≤ 30000, y ≤ 12000, x ≥ 6000, x ≥  y , x ≥ 0 and y ≥ 0 are as follows:

The corner points of the feasible region are D(6000, 0), $A\left(3000, 0\right)$, $F\left(18000, 12000\right)$ and $E\left(12000, 12000\right)$.

The values of Z at these corner points are as follows:

We see that the maximum value of the objective function Z is 246000 which is at $F\left(18000, 12000\right)$ that means at x = 18000 and y = 12000.
Thus, the optimal value of Z is 246000.
 

Answer:

First, we will convert the given inequations into equations, we obtain the following equations:
x + y = 8, x + 4y = 12, x = 3, y = 2

Region represented by x + y ≥ 8:
The line x + y = 8 meets the coordinate axes at A(8, 0) and B(0, 8) respectively. By joining these points we obtain the line x + y = 8.
Clearly (0,0) does not satisfies the inequation x + y ≥ 8. So,the region in xy plane which does not contain the origin represents the solution set of the inequation x + y ≥ 8.

Region represented by x + 4y ≥ 12:
The line x + 4y = 12 meets the coordinate axes at C(12, 0) and D(0, 3) respectively. By joining these points we obtain the line x + 4y = 12.
Clearly (0,0) satisfies the inequation x + 4y ≥ 12. So,the region in xy plane which contain the origin represents the solution set of the inequation x + 4y ≥ 12.

The line x = 3 is the line that passes through the point (3, 0) and is parallel to Y axis.x ≥ 3 is the region to the right of the line x = 3.

The line y = 2 is the line that passes through the point (0, 12) and is parallel to X axis.y ≥ 2 is the region above the line y = 2.



The corner points of the feasible region are E(3, 5) and F(6, 2).
 

The values of Z at these corner points are as follows.

Therefore, the minimum value of Z is 20 at the point F(6, 2). Hence, x = 6 and y =2 is the optimal solution of the given LPP.
Thus, the optimal value of Z is 20.

Answer:

First, we will convert the given inequations into equations, we obtain the following equations:
2x + y = 10, x + 3y = 15, x = 10, y = 8

Region represented by 2x + y ≥ 10:
The line 2x + y = 10 meets the coordinate axes at A(5, 0) and B(0, 10) respectively. By joining these points we obtain the line 2x + y = 10.
Clearly (0,0) does not satisfies the inequation 2x + y ≥ 10. So,the region in xy plane which does not contain the origin represents the solution set of the inequation 2x + y ≥ 10.

Region represented by x + 3y ≥ 15:
The line x + 3y = 15 meets the coordinate axes at C(15, 0) and D(0, 5) respectively. By joining these points we obtain the line x + 3y = 15.
Clearly (0,0) satisfies the inequation x + 3y ≥ 15. o,the region in xy plane which does not contain the origin represents the solution set of the inequation x + 3y ≥ 15.

The line x = 10 is the line that passes through the point (10, 0) and is parallel to Y axis.x ≤ 10 is the region to the left of the line x = 10.

The line y = 8 is the line that passes through the point (0, 8) and is parallel to X axis.y ≤ 8 is the region below the line y = 8.

Region represented by x ≥ 0 and y ≥ 0:
Since, every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by the inequations x ≥ 0 and y ≥ 0.

The feasible region determined by the system of constraints, 2x + y ≥ 10, x + 3y ≥ 15, x ≤ 10, y ≤ 8, x ≥ 0 and y ≥ 0 are as follows.



The corner points of the feasible region are E(3, 4),$H\left(10,\frac{5}{3}\right)$, F(10, 8) and G(1, 8).
The values of Z at these corner points are as follows.

Therefore, the minimum value of Z is 27 at the point F(3, 4). Hence, x = 3 and y =4 is the optimal solution of the given LPP.
Thus, the optimal value of Z is 27.

Answer:

First, we will convert the given inequations into equations, we obtain the following equations:
x + y = 8, x + 4y = 12, x = 0 and y = 0

5x + 8y = 20 is already an equation.

Region represented by x + y ≤ 8:
The line x + y = 8 meets the coordinate axes at A(8, 0) and B(0, 8) respectively. By joining these points we obtain the line x + y = 8.Clearly (0,0) satisfies the inequation x + y ≤ 8. So,the region in xy plane which contain the origin represents the solution set of the inequation x + y ≤ 8.

Region represented by x + 4y ≥ 12:
The line x + 4y = 12 meets the coordinate axes at C(12, 0) and D(0, 3) respectively. By joining these points we obtain the line x + 4y = 12.Clearly (0,0) satisfies the inequation x + 4y ≥ 12. So,the region in xy plane which does not contain the origin represents the solution set of the inequation x + 4y ≥ 12.

The line 5x + 8y = 20 is the line that passes through E(4, 0) and $F\left(0, \frac{5}{2}\right)$ .

Region represented by x ≥ 0 and y ≥ 0:
Since, every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by the inequations x ≥ 0 and y ≥ 0.

The feasible region determined by the system of constraints, x + y ≤ 8, x + 4y ≥ 12, 5x + 8y = 20, x ≥ 0 and y ≥ 0 are as follows.



The corner points of the feasible region are B(0,8), D(0,3), $G\left(\frac{20}{3}, \frac{4}{3}\right)$.
The values of Z at these corner points are as follows.

Therefore, the minimum value of Z is 60 at the point D(0,3).  Hence, x = 0 and y =3 is the optimal solution of the given LPP.
Thus, the optimal value of Z is 60.

Answer:

We need to maximize Z = 4x + 3y

First, we will convert the given inequations into equations, we obtain the following equations:
3x + 4y = 24, 8x + 6y = 48, x = 5 , y = 6, x = 0 and y = 0.

The line 3x + 4y = 24 meets the coordinate axis at A(8, 0) and B(0,6). Join these points to obtain the line 3x + 4y = 24.
Clearly, (0, 0) satisfies the inequation 3x + 4y ≤ 24.So, the region in xy-plane that contains the origin represents the solution set of the given equation.

The line 8x + 6y = 48 meets the coordinate axis at C(6, 0) and D(0,8). Join these points to obtain the line 8x + 6y = 48.
Clearly, (0, 0) satisfies the inequation 8x + 6y  ≤ 48. So, the region in xy-plane that contains the origin represents the solution set of the given equation.

x = 5 is the line passing through x = 5 parallel to the Y axis.
y = 6 is the line passing through y = 6 parallel to the X axis.

Region represented by x ≥ 0 and y ≥ 0:
Since, every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by the inequations.
These lines are drawn using a suitable scale.

The corner points of the feasible region are O(0, 0), $G\left(5, 0\right)$, $F\left(5,\frac{4}{3}\right)$, $E\left(\frac{24}{7},\frac{24}{7}\right)$ and $B\left(0, 6\right)$.

The values of Z at these corner points are as follows.
 

We see that the maximum value of the objective function Z is 24 which is at $F\left(5,\frac{4}{3}\right)$ and $E\left(\frac{24}{7},\frac{24}{7}\right)$.
Thus, the optimal value of Z is 24.

Answer:

First, we will convert the given inequations into equations, we obtain the following equations:
x − y = 0, − x + 2y = 2, x = 3, y = 4, x = 0 and y = 0. 

Region represented by x − y ≥ 0 or x ≥ y:
The line x − y = 0 or x = y passes through the origin.The region to the right of the line x = y will satisfy the given inequation.
Let's check by taking an example like if we take a point (4, 3) to the right of the line x = y .Here x ≥ y.So, it satisfy the given inequation.  Take a point (4, 5) to the left of the line x = y. Here, x ≤ y. That means it does not satisfy the given inequation.
 
Region represented by − x + 2y ≥ 2:
The line − x + 2y = 2 meets the coordinate axes at A(−2, 0) and B(0, 1) respectively. By joining these points we obtain the line
− x + 2y = 2.Clearly (0,0) does not satisfies the inequation − x + 2y ≥ 2. So,the region in xy plane which does not contain the origin represents the solution set of the inequation − x + 2y ≥ 2 .

The line x = 3 is the line that passes through the point (3, 0) and is parallel to Y axis. x ≥ 3 is the region to the right of the line x = 3.

The line y = 4 is the line that passes through the point (0, 4) and is parallel to X axis. y ≤ 4 is the region below the line y = 4.
 
Region represented by x ≥ 0 and y ≥ 0:
Since, every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by the inequations x ≥ 0 and y ≥ 0.

The feasible region determined by the system of constraints x − y ≥ 0,− x + 2y ≥ 2, x ≥ 3, y ≤ 4, x ≥ 0 and y ≥ 0 are as follows.



The corner points of the feasible region are $C\left(3, \frac{5}{2}\right)$,$D\left(3, 3\right)$, E(4, 4) and F(6, 4).
The values of Z at these corner points are as follows.

Therefore, the minimum value of Z is 4 at the point E(4, 4). Hence, x = 4 and y = 4 is the optimal solution of the given LPP.
Thus, the optimal value of Z is 4.

Answer:

We need to maximize Z = 3x + 5y

First, we will convert the given inequations into equations, we obtain the following equations:
x + 2y = 20, x + y = 15, y = 5 , x = 0 and y = 0.

The line x + 2y = 20 meets the coordinate axis at A(20, 0) and B(0,10). Join these points to obtain the line x + 2y = 20.
Clearly, (0, 0) satisfies the inequation x + 2y ≤ 20. So, the region in xy-plane that contains the origin represents the solution set of the given equation.

The line x + y = 15 meets the coordinate axis at C(15, 0) and D(0,15). Join these points to obtain the line x + y = 15.
Clearly, (0, 0) satisfies the inequation x + y ≤ 15. So, the region in xy-plane that contains the origin represents the solution set of the given equation.

y = 5 is the line passing through (0, 5) and parallel to the X axis.The region below the line y = 5 will satisfy the given inequation.

Region represented by x ≥ 0 and y ≥ 0:
Since, every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by the inequations.
These lines are drawn using a suitable scale.

The corner points of the feasible region are O(0, 0),$C\left(15, 0\right)$, $E\left(10, 5\right)$ and $F\left(0, 5\right)$

The values of Z at these corner points are as follows.
 

We see that the maximum value of the objective function Z is 55 which is at $E\left(10, 5\right)$.
Thus, the optimal value of Z is 55.

Answer:

First, we will convert the given inequations into equations, we obtain the following equations:
x₁ + 3x₂ = 3, x₁ + x₂ = 2, x₁ = 0 and x₂ = 0

Region represented by x₁ + 3x₂ ≥ 3 :
The line x₁ + 3x₂ = 3 meets the coordinate axes at A(3, 0) and B(0, 1) respectively. By joining these points we obtain the line x₁ + 3x₂ = 3.
Clearly (0,0) does not satisfies the inequation x₁ + 3x₂ ≥ 3 .So,the region in the plane which does not contain the origin represents the solution set of the inequation x₁ + 3x₂ ≥ 3.

Region represented by x₁ + x₂ ≥ 2:
The line x₁ + x₂ = 2 meets the coordinate axes at C(2, 0) and D(0, 2) respectively. By joining these points we obtain the line x₁ + x₂ = 2.Clearly (0,0) does not satisfies the inequation x₁ + x₂ ≥ 2. So,the region containing the origin represents the solution set of the inequation x₁ + x₂ ≥ 2.

Region represented by x₁ ≥ 0 and x₂ ≥ 0:
Since, every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by the inequations x₁ ≥ 0 and x₂ ≥ 0.

The feasible region determined by the system of constraints, x₁ + 3x₂ ≥ 3 , x₁ + x₂ ≥ 2, x₁ ≥ 0, and x₂ ≥ 0, are as follows.

The corner points of the feasible region are O(0, 0), B(0, 1), $E\left(\frac{3}{2}, \frac{1}{2}\right)$ and C(2, 0).

The values of Z at these corner points are as follows.

Therefore, the minimum value of Z is 0 at the point O(0, 0). Hence, x₁ = 0 and x₂ = 0 is the optimal solution of the given LPP.
Thus, the optimal value of Z is 0.

Answer:

First, we will convert the given inequations into equations, we obtain the following equations:
x + y = 1, 10x +y = 5, x + 10y = 1, x = 0 and y = 0

Region represented by x + y ≥ 1:
The line x + y = 1  meets the coordinate axes at A(1, 0) and B(0,1) respectively. By joining these points we obtain the line x + y = 1.
Clearly (0,0) does not satisfies the inequation x + y ≥ 1. So,the region in xy plane which does not contain the origin represents the solution set of the inequation x + y ≥ 1.

Region represented by 10x +y ≥ 5:
The line 10x +y = 5 meets the coordinate axes at $C\left(\frac{1}{2}, 0\right)$ and D(0, 5) respectively. By joining these points we obtain the line
10x +y = 5.Clearly (0,0) does not satisfies the inequation 10x +y ≥ 5. So,the region which does not contains the origin represents the solution set of the inequation 10x +y ≥ 5.

Region represented by x + 10y ≥ 1:
The line x + 10y = 1 meets the coordinate axes at $A\left(1, 0\right)$ and $F\left(0, \frac{1}{10}\right)$ respectively. By joining these points we obtain the line
x + 10y = 1.Clearly (0,0) does not satisfies the inequation x + 10y ≥ 1. So,the region which does not contains the origin represents the solution set of the inequation x + 10y ≥ 1.

Region represented by x ≥ 0 and y ≥ 0:
Since, every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by the inequations x ≥ 0, and y ≥ 0.
The feasible region determined by the system of constraints x + y ≥ 1, 10x +y ≥ 5, x + 10y ≥ 1, x ≥ 0, and y ≥ 0, are as follows.

The feasible region is unbounded.Therefore, the maximum value is infinity i.e. the solution is unbounded.

Disclaimer:
The obtained answer is for the given question. Answer in the book is 2.It would be 2 if the question is to minimize Z instead of to maximize Z.
 

Answer:

First, we will convert the given inequations into equations, we obtain the following equations:
−x₁ + 3x₂ = 10, x₁ + x₂ = 6, x₁ + x₂ = 2, x₁ = 0 and x₂ = 0

Region represented by −x₁ + 3x₂ ≤ 10:
The line −x₁ + 3x₂ = 10 meets the coordinate axes at A(−10, 0) and $B\left(0, \frac{10}{3}\right)$ respectively. By joining these points we obtain the line −x₁ + 3x₂ = 10.
Clearly (0,0) satisfies the inequation −x₁ + 3x₂ ≤ 10 .So,the region in the plane which contain the origin represents the solution set of the inequation
−x₁ + 3x₂ ≤ 10.

Region represented by x₁ + x₂ ≤ 6:
The line x₁ + x₂ = 6 meets the coordinate axes at C(6, 0) and D(0, 6) respectively. By joining these points we obtain the line x₁ + x₂ = 6.Clearly (0,0) satisfies the inequation x₁ + x₂ ≤ 6. So,the region containing the origin represents the solution set of the inequation x₁ + x₂ ≤ 6.

Region represented by x₁− x₂ ≤ 2:
The line x₁ − x₂ = 2 meets the coordinate axes at E(2, 0) and F(0, −2) respectively. By joining these points we obtain the line x₁ − x₂ = 2.Clearly (0,0) satisfies the inequation x₁− x₂ ≤ 2. So,the region containing the origin represents the solution set of the inequation x₁− x₂ ≤ 2.

Region represented by x₁ ≥ 0 and x₂ ≥ 0:
Since, every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by the inequations x₁ ≥ 0 and x₂ ≥ 0.

The feasible region determined by the system of constraints, −x₁ + 3x₂ ≤ 10, x₁ + x₂ ≤ 6, x₁− x₂ ≤ 2, x₁ ≥ 0, and x₂ ≥ 0, are as follows.

The corner points of the feasible region are O(0, 0), E(2, 0), H(4, 2), G(2, 4) and $B\left(0, \frac{10}{3}\right)$.

The values of Z at these corner points are as follows.

We see that the maximum value of the objective function Z is $\frac{20}{3}$ which is at $B\left(0, \frac{10}{3}\right)$.

Answer:

We need to maximize Z = x + y

First, we will convert the given inequations into equations, we obtain the following equations:
−2x + y = 1, x = 2, x + y = 3, x = 0 and y = 0.

The line −2x + y = 1 meets the coordinate axis at $A\left(\frac{-1}{2}, 0\right)$ and B(0, 1). Join these points to obtain the line −2x + y = 1 .
Clearly, (0, 0) satisfies the inequation −2x + y ≤ 1. So, the region in xy-plane that contains the origin represents the solution set of the given equation.

x = 2 is the line passing through (2, 0) and parallel to the Y axis.The region below the line x = 2 will satisfy the given inequation.

The line x + y = 3 meets the coordinate axis at C(3, 0) and D(0, 3). Join these points to obtain the line x + y = 3.
Clearly, (0, 0) satisfies the inequation x + y ≤ 3. So, the region in xy-plane that contains the origin represents the solution set of the given equation.

Region represented by x ≥ 0 and y ≥ 0:
Since, every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by the inequations.
These lines are drawn using a suitable scale.

The corner points of the feasible region are O(0, 0),$G\left(2, 0\right)$, $E\left(2, 1\right)$ and $F\left(\frac{2}{3}, \frac{7}{3}\right)$

The values of Z at these corner points are as follows.
 

We see that the maximum value of the objective function Z is 3 which is at $E\left(2, 1\right)$ and $F\left(\frac{2}{3}, \frac{7}{3}\right)$.
Thus, the optimal value of Z is 3.

Answer:

First, we will convert the given inequations into equations, we obtain the following equations:
x₁ − x₂ = −1, −x₁ + x₂ = 0, x₁ = 0 and x₂ = 0

Region represented by x₁ − x₂ ≤ −1:
The line x₁ − x₂ = −1 meets the coordinate axes at A(−1, 0) and B(0, 1) respectively. By joining these points we obtain the line x₁ − x₂ = −1.
Clearly (0,0) does not satisfies the inequation x₁ − x₂ ≤ −1 .So,the region in the plane which does not contain the origin represents the solution set of the inequation x₁ − x₂ ≤ −1.

Region represented by −x₁ + x₂ ≤ 0 or x₁ ≥ x₂:
The line −x₁ + x₂ = 0 or x₁ = x₂ is the line passing through (0, 0).The region to the right of the line x₁ = x₂ will satisfy the given inequation −x₁ + x₂ ≤ 0.
If we take a point (1, 3) to the left of the line x₁ = x₂. Here, 1≤3 which is not satifying the inequation x₁ ≥ x_2. Therefore, region to the right of the line x₁ = x₂ will satisfy the given inequation −x₁ + x₂ ≤ 0.

Region represented by x₁ ≥ 0 and x₂ ≥ 0:
Since, every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by the inequations x₁ ≥ 0 and x₂ ≥ 0.

The feasible region determined by the system of constraints, x₁ − x₂ ≤ −1, −x₁ + x₂ ≤ 0, x₁ ≥ 0, and x₂ ≥ 0, are as follows.

We observe that the feasible region of the given LPP does not exist.
 

Answer:

First, we will convert the given inequations into equations, we obtain the following equations:
x − y = 1, x + y = 3, x = 0 and y = 0

Region represented by x − y ≤ 1:
The line x − y = 1 meets the coordinate axes at A(1, 0) and B(0, −1) respectively. By joining these points we obtain the line x − y = 1.
Clearly (0,0) satisfies the inequation x + y ≤ 8. So,the region in xy plane which contain the origin represents the solution set of the inequation x − y ≤ 1.

Region represented by x + y ≥ 3:
The line x + y = 3 meets the coordinate axes at C(3, 0) and D(0, 3) respectively. By joining these points we obtain the line x + y = 3.
Clearly (0,0) satisfies the inequation x + y ≥ 3. So,the region in xy plane which does not contain the origin represents the solution set of the inequation x + y ≥ 3.


Region represented by x ≥ 0 and y ≥ 0:
Since, every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by the inequations x ≥ 0 and y ≥ 0.

The feasible region determined by the system of constraints x − y ≤ 1, x + y ≥ 3, x ≥ 0 and y ≥ 0 are as follows.



The feasible region is unbounded.We would obtain the maximum value at infinity.
Therefore,maximum value will be infinity i.e. the solution is unboundedm

Answer:

First, we will convert the given inequations into equations, we obtain the following equations:
5x + y = 10, x +y = 6, x + 4y = 12, x = 0 and y = 0

Region represented by 5x + y ≥ 10:
The line 5x + y = 10  meets the coordinate axes at A(2, 0) and B(0, 10) respectively. By joining these points we obtain the line 5x + y = 10.
Clearly (0,0) does not satisfies the inequation 5x + y ≥ 10. So,the region in xy plane which does not contain the origin represents the solution set of the inequation 5x + y ≥ 10.

Region represented by x +y  ≥ 6:
The line x +y = 6 meets the coordinate axes at C(6,0) and D(0, 6) respectively. By joining these points we obtain the line
2x +3y = 30.Clearly (0,0) does not satisfies the inequation x +y  ≥ 6. So,the region which does not contain the origin represents the solution set of the inequation 2x +3y ≥ 30.

Region represented by x + 4y ≥ 12
The line x + 4y = 12 meets the coordinate axes at E(12, 0) and F(0, 3) respectively. By joining these points we obtain the line
x + 4y = 12.Clearly (0,0) does not satisfies the inequation x + 4y ≥ 12. So,the region which does not contain the origin represents the solution set of the inequation x + 4y ≥  12.

Region represented by x ≥ 0 and y ≥ 0:
Since, every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by the inequations x ≥ 0, and y ≥ 0.
The feasible region determined by the system of constraints 5x + y ≥ 10, x +y  ≥ 6,x + 4y ≥ 12, x ≥ 0, and y ≥ 0, are as follows.

The corner points of the feasible region are B(0, 10), G(1,5), H(4,2) and E(12,0).

The values of Z at these corner points are as follows.

Therefore, the minimum value of Z is 13 at the point G(1,5). Hence, x = 1 and y = 5 is the optimal solution of the given LPP.
Thus, the optimal value of Z is 13.

Answer:

First, we will convert the given inequations into equations, we obtain the following equations:
x + y = 4, x + y = 3, x − 2y = 2, x = 0 and y = 0.

The line x + 3y = 6 meets the coordinate axis at $A\left(6, 0\right)$ and B(0, 2). Join these points to obtain the line x + 3y = 6.
Clearly, (0, 0) does not satisfies the inequation x + 3y ≥ 6. So, the region in xy-plane that does not contains the origin represents the solution set of the given equation.

The line x − 3y = 3 meets the coordinate axis at C(3, 0) and D(0, −1). Join these points to obtain the line x − 3y = 3.
Clearly, (0, 0) satisfies the inequation x − 3y ≤ 3. So, the region in xy-plane that contains the origin represents the solution set of the given equation.

The line 3x + 4y = 24 meets the coordinate axis at E(8, 0) and F(0, 6). Join these points to obtain the line 3x + 4y = 24.
Clearly, (0, 0) satisfies the inequation 3x + 4y ≤ 24. So, the region in xy-plane that contains the origin represents the solution set of the given equation.

The line −3x + 2y = 6 meets the coordinate axis at G(−2, 0) and H(0, 3). Join these points to obtain the line −3x + 2y = 6.
Clearly, (0, 0) satisfies the inequation −3x + 2y ≤ 6. So, the region in xy-plane that contains the origin represents the solution set of the given equation.

The line 5x + y = 5 meets the coordinate axis at $I\left(1, 0\right)$ and J(0, 5). Join these points to obtain the line 5x + y = 5.
Clearly, (0, 0) does not satisfies the inequation 5x + y ≥ 5. So, the region in xy-plane that does not contains the origin represents the solution set of the given equation.

Region represented by x ≥ 0 and y ≥ 0:
Since, every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by the inequations.
These lines are drawn using a suitable scale.

The corner points of the feasible region are P$\left(\frac{4}{13},\frac{45}{13} \right)$, K$\left(\frac{4}{3}, 5 \right)$, L$\left(\frac{84}{13},\frac{15}{13} \right)$, M$\left(\frac{9}{2},\frac{1}{2} \right)$, N$\left(\frac{9}{14},\frac{25}{14} \right)$

The values of Z at these corner points are as follows.
 

We see that the minimum value of the objective function Z is $\frac{43}{14}$ which is at N$\left(\frac{9}{14},\frac{25}{14} \right)$ and maximum value of the objective function is $\frac{183}{13}$ which is at L$\left(\frac{84}{13},\frac{15}{13} \right)$.

 

Answer:

First, we will convert the given inequations into equations, we obtain the following equations:
−2x + y = 4, x + y = 3, x − 2y = 2, x = 0 and y = 0.

The line −2x + y = 4 meets the coordinate axis at $A\left(-2, 0\right)$ and B(0, 4). Join these points to obtain the line −2x + y = 4.
Clearly, (0, 0) satisfies the inequation −2x + y ≤ 4. So, the region in xy-plane that contains the origin represents the solution set of the given equation.

The line x + y = 3 meets the coordinate axis at C(3, 0) and D(0, 3). Join these points to obtain the line x + y = 3.
Clearly, (0, 0) does not satisfies the inequation x + y ≥ 3. So, the region in xy-plane that does not contains the origin represents the solution set of the given equation.

The line x − 2y = 2 meets the coordinate axis at E(2, 0) and F(0, −1). Join these points to obtain the line x − 2y = 2.
Clearly, (0, 0) satisfies the inequation x − 2y ≤ 2. So, the region in xy-plane that contains the origin represents the solution set of the given equation.

Region represented by x ≥ 0 and y ≥ 0:
Since, every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by the inequations.
These lines are drawn using a suitable scale.

The corner points of the feasible region are $B\left(0, 4\right)$, $D\left(0, 3\right)$ and $G\left(\frac{8}{3}, \frac{1}{3}\right)$

The values of Z at these corner points are as follows.
 

We see that the minimum value of the objective function Z is $\frac{29}{3}$ which is at $G\left(\frac{8}{3}, \frac{1}{3}\right)$.
Thus, the optimal value of Z is $\frac{29}{3}$.