Maths 4 Marks

Let x units of first product and y units of second product be manufactured.

Therefore, x, y ≥ 0

The given information can be tabulated as follows:

Therefore, the constraints are

$2\text{x}+4\text{y}\leq20$

$2\text{x}+2\text{y}\leq12$

$4\text{x}+0\text{y}\leq16$

$4\text{x}\leq16$

It is known that the first product gives a profit of 2 monetary units per unit and the second 3.

Therefore, profit gained from x units of first product and y units of second product is 2x monetary units and 4y monetary units respectively.

Total profit = Z = 2x + 3y which is to be maximised

Thus, the mathematical formulat​ion of the given linear programmimg problem is 

Max  Z =  2x + 3y

Subject to

$2\text{x}+4\text{y}\leq20$

$2\text{x}+2\text{y}\leq12$

$4\text{x}+0\text{y}\leq16$

$4\text{x}\leq16$

$\text{x},\text{y}\geq0$

First we will convert inequations into equations as follows:

2x + 4y = 20, 2x + 2y = 12, 4x = 16, x = 0 and y = 0

Region represented by 2x + 4y ≤ 20:

The line 2x + 4y = 20 meets the coordinate axes at A₁(10, 0) and B₁(0, 5) respectively.

By joining these points we obtain the line 2x + 4y = 20.

Clearly (0, 0) satisfies the 3x + 2y = 210.

So, the region which contains the origin represents the solution set of the inequation 2x + 4y ≤ 20.

Region represented by 2x + 2y ≤ 12:

The line 2x + 2y = 16 meets the coordinate axes at C₁(6, 0) and D₁(0, 6) respectively.

By joining these points we obtain the line 2x + 2y = 12.

Clearly (0, 0) satisfies the inequation 2x + 2y ≤ 12.

So, the region which contains the origin represents the solution set of the inequation 2x + 2y ≤ 12.

Region represented by 4x ≤ 16:

The line 4x =16 or x = 4 is the line passing through the point E₁(4, 0) and is parallel to Y axis.

The region to the left of the line x = 4 would satisfy the inequation 4x ≤ 16.

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 + 4y ≤ 20, 2x + 2y ≤ 12, 4x ≤ 16, x ≥ 0 and y ≥ 0 are as follows.

The corner points are O(0, 0), B₁(0, 5), G₁(2, 4), F₁(4, 2) and E₁(4, 0).

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

The maximum value of Z is 16 which is attained at G₁(2, 4)

Suppose tailor A and B work for x and y days respectively.

Since, tailor A and B earn Rs. 15 and Rs. 20 respectively.

So, tailor A and B earn is X and Y days Rs. 15x and 20y respectively, let z denote maximum profit that gives minimum labour cost, so,

Z = 15x + 20y

Since, Tailor A and B stitch 6 and 10 shirts respectively in a day, so, tailor A can stitch 6x and B can stitch 10y shirts in x and y days respectively, but it is desired to produce 60 shirts at least, so

$6\text{x}+10\text{y}\geq60$

$3\text{x}+5\text{y}\geq30$ (first constraint)

Since, Tailor A and B stitch 4 pants per day each, so, tailor A can stitch 4x and B can stitch 4y pants in x and y days respectively, but it is desired to produce at least 32 pants, so

$4\text{x}+4\text{y}\geq32$

$\text{x}+\text{y}\geq8$ (second constraint)

Hence, mathematical formulation of LPP is,

Find x and y which minimize

Z = 15x + 20y

Subject to constraints,

$3\text{x}+5\text{y}\geq30$

$\text{x}+\text{y}\geq8$

$\text{x},\text{y}\geq0$ [Since x and y not be less than zero]

Region $3\text{x}+5\text{y}\geq30$:

Line 3x + 5y = 30 meets axes at A₁(10, 0), B₁(0, 6) respectively.

Region not containing origin represents $3\text{x}+5\text{y}\geq30$ as (0, 0) does not satisfy $3\text{x}+5\text{y}\geq30$.

Region $\text{x}+\text{y}\geq8$:

Line x + y = 8 meets axes at A₂(8, 0), B₂(0, 8) respectively.

Region not containing origin represents $\text{x}+\text{y}\geq8$ as (0, 0) does not satisfy $\text{x}+\text{y}\geq8$.

Region $\text{x},\text{y}\geq0$:

It represent first quadrant.

Unbounded shaded region AP₁B₂ represents feasible region with corner points A₁(10, 0), P(5, 3),B₂(0, 8).

The value of Z = 15x + 20y at

A₁(10, 0) = 15(10) + 20(0) = 150

P(5, 3) = 15(5) + 20(3) = 135

B₂(0, 8) = 15(0) + 20(8) = 160

Smallest value of Z is 135,

Now open half plane 15x + 20y < 135 has no point in common with feasible region, so smallest value is the minimum value.

So, Z = 135, at x = 5, y = 3

Tailor A should work for 5 days and B should work for 3 days.

Let x bottles of medicine A and y bottles of medicine B are prepared. Number of bottles cannot be negative.

Converting the given inequations in to equations.

3x + y = 12, 2x + 5y = 34, x = y = 0

Region represented by $3\text{x}+\text{y}\leq12$:

Line 3x + y = 12 meets the coordinate axes at A₁(4, 0) and B(0, 12), clearly, (0, 0) satisfies $3\text{x}+\text{y}\leq12$, so, region containing origin is represented by $3\text{x}+\text{y}\leq12$ in xy-plane.

Region represented by $2\text{x}+5\text{y}\leq34$:

Line 2x +y = 34 meets coordinate axes at A₂ (17, 0) and $\text{B}\Big(0,\frac{34}{5}\Big)$ clearly, (0, 0) satisfies the $2\text{x}+5\text{y}\leq34$ so, region containing origin represents $2\text{x}+5\text{y}\leq34$ in xy-plane.

Region represented by $\text{x},\text{y}\geq0$:

It represent the first quadrant in xy-plane

Therefore, shaded area OA₁PB₂ is the feasible region.

The coordinate of P(2, 6) is obtained by solving 2x + 5y = 34 and 3x + y = 12

The value of Z = 10x + 6y at

$\text{O}(0, 0) = 10(0) + 6(0) = 0$

$\text{A}_1(4, 0) = 10(4) + 6(0) = 40$

$\text{P}(2, 6) = 10(2) + 6(6) = 56$

$\text{B}_2\Big(0,\frac{34}{5}\Big)=10(0)+6\Big(\frac{34}{5}\Big)=\frac{204}{5}=40\frac{4}{5}$

Hence, maximum Z = 56 at x = 2, y = 6.

Let the manufacturer produces x number of models X and y number of model Y bikes. Model X takes 6 man-hours to make per unit and model Y takes 10 man-hours to make per unit.

There is total of 450 man-hours available per week.

$\therefore6\text{x}+10\text{y}\leq450$

$\Rightarrow3\text{x}+5\text{y}\leq225\ .....(\text{i})$

For model X and Y, handling and marketing costs are Rs. 2000 and Rs. 1000, respectively, total funds available for these purposes are Rs. 80000 per week.

$\therefore2000\text{x}+1000\text{y}\leq80000$

$\Rightarrow2\text{x}+\text{y}\leq80\ .....(\text{ii})$

Also $\text{x}\geq0,\text{y}\geq0$

the profits per unit for models X and Y are Rs. 1000 and Rs. 500, respectively,

So, we have to maximize Z - 1000x + 500y Subject to $3\text{x}+5\text{y}\leq225,2\text{x}+\text{y}\leq80,\text{x}\geq0,\text{y}\geq0$

These inequalities are plotted as shown in the figure.

So, for maximum profit manufacturer must produces 25 number of model X and 30 number of model bikes.

Let x units of chairs and y units of tables were produced

Therefore, $\text{x},\text{y}\geq0$

The given information can be tabulated as follows:

Therefore, the constraints are

$5\text{x}+20\text{y}\leq400$

$10\text{x}+25\text{y}\leq450$

It is known that to make a chair requires 5 square feet of wood and 10 man-hours and yields a profit of Rs. 45, while each table uses 20 square feet of wood and 25 man-hours and yields a profit of Rs. 80.

Therefore, profit gained to make x chairs and y tables is Rs. 45x and Rs. 80y respectively Total profit = Z = 45x +80y which is to be maximised.

Thus, the mathematical formulation of the given linear programming problem is

Max 2 = 45x + 80y

Subject to

$5\text{x}+20\text{y}\leq400$

$10\text{x}+25\text{y}\leq450$

$\text{x},\text{y}\geq0$ 

First we will convert inequations into equations as follows:

5x + 20y = 400, 10x + 25y = 450, X = 0 and y = 0

Region represented by $5\text{x}+20\text{y}\leq400$:

The line 5x + 20y = 400 meets the coordinate axes at A(80, 0) and B(0, 20) respectively.

By joining these points we obtain the line 5x + 20y = 400.

Clearly (0, 0) satisfies the $5\text{x}+20\text{y}\leq400$.

So, the region which contains the origin represents the solution set of the inequation $5\text{x}+20\text{y}\leq400$.

Region represented by $10\text{x}+25\text{y}\leq450$:

The line 10x + 25y = 450 meets the coordinate axes at C(45, 0) and D(0, 18) respectively.

By joining these points we obtain the line 10x + 25y = 450.

Clearly (0, 0) satisfies the inequation $10\text{x}+25\text{y}\leq450$.

So the region which contains the origin represents the solution set of the inequation $10\text{x}+25\text{y}\leq450$.

Region represented by $\text{x}\geq0$ and $\text{y}\geq0$:

Since, every point in the first quadrant satisfies these inequations.

So, the first quadrant is the region represented by the inequations $\text{x}\geq0$, and $\text{y}\geq0$.

The feasible region determined by the system of constraints $5\text{x}+20\text{y}\leq400,10\text{x}+25\text{y}\leq450,\text{x}\geq0,$  and $\text{y}\geq0$ are as follows.

The corner points are A(0, 18), B(45, 0)

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

The maximum value of Z is 2025 which is attained at B(45, 0).

Thus, the maximum profit is of Rs. 2025 obtained when 45 units of chairs and no units of tables are produced.

Let x and y be the number of packets of food P and Q respectively, $\text{x}\ge0,\ \text{y}\ge0.$

We have to maximize Z = 6x + 3y (vitamin A) subject to the constraints $12\text{x}+3\text{y}\ge240$ (constraints on Calcium), $\text{i.e., }4\text{x}+\text{y}\ge80\dots(\text{i})$

$\text{And }4\text{x}+50\text{y}\ge460$ (constraints on Iron), $\text{i.e., }\text{x}+5\text{y}\ge115\dots(\text{ii})$

$\text{Also }6\text{x}+4\text{y}\le300$ (constraints on Cholesterol), i.e., $3\text{x}+2\text{y}\le150\dots(\text{iii})$

$\text{x}\ge0,\ \text{y}\ge0\dots(\text{iv})$

Consider $4\text{x}+\text{y}\ge80$

Let 4x + y = 80

⇒ y = 80 - 4x

Here, (0, 0) does not satisfy this inequation, therefore the required half plane does not include the point (0, 0)

Again consider $\text{x}+5\text{y}\ge115$

Let x + 5y = 115

⇒ x = 115 - 5y

Here, also (0, 0) does not satisfy this inequation, therefore the required half plane does not include the point (0, 0)

Again consider $3\text{x}+2\text{y}\le150$

$\text{Let }3\text{x}+2\text{y}=150\Rightarrow\frac{\text{x}}{50}+\frac{\text{y}}{75}=1$

Therefore, G(50, 0) and H(0, 75) satisfy the equation.

As (0, 0) satisfies the inequation 3x + 2y = 150, therefore the required half plane contains (0, 0). The shaded region is the feasible solution and its corners are P(15, 20), Q(40, 15) and R(2, 72).

Hence, maximum Z = 285 units of vitamin A at x = 40, y = 15.

The company makes profit of Rs. 12 and Rs. 16 per doll respectively on doll A and doll B

Step I

= Z = 12x + 16y

The production level of x + y should not exceed 1200

x + y ≤ 1200

The production level of dolls of type A exceeds three times the production of dolls of type B by at most 600

x - 3y ≤ 600

Demand for dolls of type B is at most of that for dolls of type A.

$\text{y}\leq\frac{\text{x}}{2}$

Hence we can say the LPP is subject to maximize Z = 12x + 16y and to constraints

$\text{x}+\text{y}\leq1200$

$\text{x}-3\text{y}\leq600$

$\text{y}\leq\frac{\text{x}}{2}$

$\text{x},\text{y}\geq0$

Step II

1. The line x + y = 1200 passes through A(1200, 0),B(0, 1200)

Put x = 0, y = 0 in x + y ≤ 1200

We get 0 ≤ 1200 which is true.

$\therefore$ x + y ≤ 1200 lies on and below AB

2. The line x - 3y = 600, passes through C(600, 0),D(0, –200)

Put x = 0,y = 0 in x - 3y 5600

Clearly 0 ≤ 600 is true.

Hence x - 3y ≤ 600 lies above the line CDCD

3. $\text{y}\leq\frac{\text{x}}{2}$ passes through (400, 200) and (0, 0)

Put x = 200, y = 0 in $\text{y}\leq\frac{\text{x}}{2}$

$0\text{}\leq\frac{\text{200}}{2}$ is also true.

→ (200, 0) lies in the region (i.e) below OP

4. x ≥ 0 lies on and to the right of y-axis.
5. y ≥ 0 lies on and above X-axis.

The shaded portion PQCO represents the feasible region.

Step III

The point P is in the intersection of the lines x + y = 1200,

$\text{y}\leq\frac{\text{x}}{2}$

⇒ 2y = x

On solving these two equations we get,

x = 800 and y = 400

The point Q is in the intersection of the lines x + y = 1200 and x - 3y = 600

Let us solve these two equations to get the value of x and y

x + y = 1200

x - 3y = 600

4y = 600

y = 150

Hence x = 1050

The coordinates of Q are (1050, 150).

The point C is (600, 0)

The objective function is Z = 12x + 16y

Step IV

Let us now obtain the values of objective function as follows:

At the points (x, y) the value of the objective function Z = 12x + 16

At P(800, 400) the value of the objective function

Z = 12 × 800 + 16 × 400 = 16,000

At Q(1050, 150) the value of the objective function

Z = 12 × 1050 + 16 × 150 = 15,000

At C(600, 0) the value of the objective function Z = 12 × 600 + 16 × 0 = 7200

At 0(0, 0) the value of the objective function Z =12 × 0 + 16 × 0 = 0

It is clear that Z is maximum at P(800, 400).

The maximum value of Z is Rs. 16,000.

Thus to maximize the profit 800 dolls of type A and 400 dolls of type B should be produced to get a maximum profit of Rs. 16,000.

Let required quantity of food A and B be x and y units respectively.

Costs of one unit of food A and B are Rs. 4 and Rs. 3 per unit respectively, so, costs of x unit of food A and y unit of food B are 4x and 3y respectively.

Let Z be minimum total cost, so

Z = 4x + 3y

Since one unit of food A and B contain 200 and 100 units of vitamin respectively.

So, x units of food A and y units of food B contain 200x and 100y units of vitamin but minimum requirement of vitamin is 4000 units, so

$200\text{x}+100\text{y}\geq4000$ 

$\Rightarrow2\text{x}+\text{y}\geq40$ (first constraint)

Since one unit of food A and B contain 1 unit and 2 unit of minerals, so x units of food A and y units of food B contain x and 2y units of minerals respectively but minimum requirement of minerals is 50 units, so

$\text{x}+2\text{y}\geq50$ (second constraint)

Since one unit of food A and B contain 40 calories each, so x units of food A and y units of food B contain 40x and 40y calories respectively but minimum requirement of calories is 1400, so

$40\text{x}+40\text{y}\geq1400$

$\Rightarrow2\text{x}+2\text{y}\geq70$

$\Rightarrow\text{x}+\text{y}\geq35$ (third constraint)

So, mathematical formulation of LPP is find x and y which

minimize Z = 4x + 3y

Subject to constraint,

 $2\text{x}+\text{y}\geq40$

 $\text{x}+2\text{y}\geq50$

 $\text{x}+\text{y}\geq35$

$\text{x},\text{y}\geq0$ [Since quantity of food can not be less than zero]

Region $2\text{x}+\text{y}\geq40$:

Line 2x + y = 40 meets axes at A₁(20, 0), B₁(0, 40) region not containing origin represents 2x +y > 40 as (0, 0) does not satisfy $2\text{x}+\text{y}\geq40$.

Region $\text{x}+2\text{y}\geq50$:

Line x + 2y = 50 meets axes at A₂(50, 0), B₂(0, 25).

Region not containing origin represents $\text{x}+2\text{y}\geq50$ as (0, 0) does not satisfy x + 2y = 50.

Region $\text{x}+\text{y}\geq35$:

Line x + y = 35 meets axes at A₃ (35, 0), B₃(0, 35).

Region not containing origin represents $\text{x}+\text{y}\geq35$ as (0, 0) does not satisfy $\text{x}+\text{y}\geq35$.

Region $\text{x},\text{y}\geq0$:

It represent first quadrant in xy-plane.

Unbounded shaded region A₂PQB₁ represents feasible region with corner points A₂(50, 0), P(20,15), Q(5, 30), B₂(0, 40)

The value of Z = 4x + 3y at

A₂(50, 0) = 4(50) + 3(0) = 2000

P(20, 15) = 4(20) + 3(15) = 125

Q(5, 30) = 4(5) + 3(30) = 110

B₁(0, 40) = 4(0) + 3(40) = 110

Smallest value of Z = 110

Open half plane $4\text{x}+3\text{y}\leq110$ has no point in common with feasible region, so, smallest value is the minimum value.

Hence,

quantity of food A = x = 5 unit

quantity of food B = y = 30 unit

minimum cost = Rs 110.

Since, per unit of food I costs Rs. 0.60 and that of food II costs Rs. 1.00.

Therefore, x lbs of food I costs Rs. 0.60 x and y lbs of food II costs Rs. 1.00y.

Total cost per day = Rs. (0.60x + 1.00y)

​Let Z denote the total cost per day

Then, Z = 0.60x + 1.00y

Since, each unit of food I contains 10 units of calcium.

Therefore, x units of food I contains 10x units of calcium.

Each unit of food II contains 4 units of calcium.

So, y units of food II contains 4y units of calcium.

Thus, x units of food I and y units of food II contains (10x + 4y) units of calcium.

But, the minimum requirement is 20 units of calcium.

$\therefore10\text{x}+4\text{y}\geq20$

Since, each unit of food I contains 5 units of protein.

Therefore, x units of food I contains 5x units of protein.

Each unit of food II contains 6 units of protein.

So, y units of food II contains 6y units of protein.

Thus, x units of food I and y units of food II contains (5x + 6y) units of protein.

But, the minimum requirement is 20 lbs of protein.

$\therefore5\text{x}+6\text{y}\geq20$

Since, each unit of food I contains 2 units of calories.

Therefore, x units of food I contains 2x units of calories.

Each unit of food II contains 6 units of calories.

So, y units of food II contains 6y units of calories.

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

But, the minimum requirement is 12 lbs of calories.

$\therefore2\text{x}+6\text{y}\geq12$

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

So, $\text{x},\text{y}\geq0$

Hence, the required LPP is as follow:

Min Z = 0.66x + 1.00y

Subject to

$10\text{x}+4\text{y}\geq20$

$5\text{x}+6\text{y}\geq20$

$2\text{x}+6\text{y}\geq12$

$\text{x},\text{y}\geq0$

First, we will convert the given inequations into equations, we obtain the following equations:

10x + 4y = 20, 5x +6y = 20, 2x + 6y = 12, x = 0 and y = 0

Region represented by 10x + 4y ≥ 20:

The line 10x + 4y = 20 meets the coordinate axes at A(2, 0) and B(0, 5) respectively.

By joining these points we obtain the line 10x + 4y = 20.

Clearly (0, 0) does not satisfies the inequation 10x + 4y ≥ 20.

So, the region in xy plane which does not contain the origin represents the solution set of the inequation 10x + 4y ≥ 20.

Region represented by $5\text{x}+6\text{y}\geq20$:

The line 5x + 6y = 20 meets the coordinate axes at C(4, 0) and $\text{D}\Big(0,\frac{10}{3}\Big)$ respectively.

By joining these points we obtain the line 5x + 6y = 20.

Clearly (0, 0) does not satisfies the inequation $5\text{x}+6\text{y}\geq20$.

So, the region in xy plane which does not contain the origin represents the solution set of the inequation $5\text{x}+6\text{y}\geq20$.

Region represented by 2x + 6y ≥ 12:

The line 2x + 6y =12 meets the coordinate axes at E(6, 0) and F(0, 2) respectively.

By joining these points we obtain the line 2x + 6y =12.

Clearly (0, 0) does not satisfies the inequation 2x + 6y ≥ 12.

So, the region which does not contains the origin represents the solution set of the inequation 2x + 6y ≥ 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 10x + 4y ≥ 20, 5x +6y ≥ 20, 2x + 6y ≥ 12, x ≥ 0, and y ≥ 0 are as follows.

The set of all feasible solutions of the above LPP is represented by the feasible region shaded in the graph.

The corner points of the feasible region are B(0, 5), $\text{G}\Big(1,\frac{5}{2}\Big),\text{H}\Big(\frac{8}{3},\frac{10}{9}\Big)$ and E(6, 0).

The value of the objective function at these points are given by the following table:

We see that the minimum cost is 2.7 which is at $\Big(\frac{8}{3},\frac{10}{9}\Big)$.

Thus, at minimum cost, $\frac{8}{3}$ unit of food I and $\frac{10}{9}$ units of food II should be included in the diet.

The given contraints are:

$4\text{x}+\text{y}\geq80$

$\text{x}+5\text{y}\geq115$

$3\text{x}+2\text{y}\leq150$

$\text{x}\geq0,\text{y}\geq0$

Converting the given inequation into equation, we get

4x + y = 80, x + 5y = 115, 3x + 2y = 150, x = 0 and y = 0

These lines are drawn on the graph and the shaded region ABC represents the feasible region of the given LPP.

It can be observed that the feasible region is bounded.

The coordinates of the corner points of the feasible region are A(2, 72), B(15, 20), and C(40, 15).

The values of the objective function, Z at these corner points are given in the following table:

From the table, Z is manimum at x = 15 and y = 20 and the manimum value of Z is 150.

Thus, the manimum value of Z is 150.

The given contraints are:

$\text{x}+2\text{y}\leq28$

$3\text{x}+\text{y}\leq24$

$\text{x}\geq2$

$\text{x},\text{y}\geq0$

Converting the given inequation into equation, we get

x + 2y = 28, 3x + y = 24, x = 2 and y = 0

These lines are drawn on the graph and the shaded region ABCD represents the feasible region of the given LPP.

It can be observed that the feasible region is bounded.

The coordinates of the corner points of the feasible region are A(2, 13), B(2, 0), C(4, 12) and D(8, 0).

The values of the objective function, Z at these corner points are given in the following table:

From the table, Z is maximum at x = 4 and y = 12 and the maximum value of Z is 200.

Thus, the maximum value of Z is 200.

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(6, 0) and B(0, 2).

Join these points to obtain the line x + 3y = 6.

Clearly, (0, 0) does not satisfies the inequation $\text{x}+3\text{y}\geq6$.

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 $\text{x}-3\text{y}\leq3$.

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 $3\text{x}+4\text{y}\leq24$.

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 $-3\text{x}+2\text{y}\leq6$.

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(1, 0) and J(0, 5).

Join these points to obtain the line 5x + y = 5.

Clearly, (0, 0) does not satisfies the inequation $5\text{x}+\text{y}\geq5$.

So, the region in xy-plane that does not contains the origin represents the solution set of the given equation.

Region represented by $\text{x}\geq0$ and $\text{y}\geq0$:

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.

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 $\text{x}_1-\text{x}_2\leq-1$:

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 $\text{x}_1-\text{x}_2\leq-1$.

So,the region in the plane which does not contain the origin represents the solution set of the inequation $\text{x}_1-\text{x}_2\leq-1$.

Region represented by $-\text{x}_1+\text{x}_2\leq0$ or $\text{x}_1\geq\text{x}_2$:

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 $-\text{x}_1+\text{x}_2\leq0$.

If we take a point (1, 3) to the left of the line x₁ = x₂.

Here, $1\leq3$  which is not satifying the inequation $\text{x}_1\geq\text{x}_2$.

Therefore, region to the right of the line x₁ = x₂ will satisfy the given inequation $-\text{x}_1+\text{x}_2\leq0$.

Region represented by $\text{x}_1\geq0$ and $\text{x}_2\geq0$:

Since, every point in the first quadrant satisfies these inequations.

So, the first quadrant is the region represented by the inequations $\text{x}_1\geq0$ and $\text{x}_2\geq0$.

The feasible region determined by the system of constraints, $\text{x}_1-\text{x}_2\leq-1,-\text{x}_1+\text{x}_2\leq0,\text{x}_1\geq0$ and $\text{x}_2\geq0$, are as follows.

Region represented by $5\text{x}+\text{y}\geq10$:

Line 5x + y - 10 meets coordinate axes at A₁(2, 0) and B₁(0, 10).

Clearly, (0, 0) does not satisfy $5\text{x}+\text{y}\geq10$, so region not containing origin represents $5\text{x}+\text{y}\geq10$ in xy -plane.

Region represented by $\text{x}+\text{y}\geq6$:

Line x + y = 6 meets coordinate axes at A₂(6, 0) and B₂(0, 6).

Clearly, (0, 0) does not satisfy $\text{x}+\text{y}\geq6$, so region not containing origin represents $\text{x}+\text{y}\geq6$ in xy -plane.

Region represented by $\text{x}+4\text{y}\geq12$:

Line x + 4y = 12 meets coordinate axes at A₃(12, 0) and B₃(0, 3).

Clearly, (0, 0) does not satisfy $\text{x}+4\text{y}\geq12$, so, region not containing origin $\text{x}+4\text{y}\geq12$ in xy - plane.

Region represented by $\text{x}\geq,\text{y}\geq0$:

It represents first quadrant in xy-plane.

The unbounded shaded region with corner points A₃(12, 0), P(4, 2), Q(1, 5), B₁(0, 10) represents feasible region.

Point P is obtained by solving x + 4y = 12 and x + y = 6, Q by solving x + y = 6 and 5x + y = 10.

The value of Z = 3x + 2y at

A₃(12, 0) = 3(12) + 2(0) = 36

P(4, 2) = 3(4) + 2(2) = 16

Q(1, 5) = 3(1) + 2(5) = 13

B(0, 10) = 3(0) + 2(10) = 20

Smallest value of Z = 13,

Now open half plane $3\text{x}+2\text{y}\leq13$ has no point in common with feasible region, so, smallest value is the minimum value of Z,

Hence,

Minimum z = 13 at x = 1, y = 5

We have to maximize Z = 60x + 15y.

First, we will convert the given inequations into equations, we obtain the following equations:

x + y = 50, 3x + y = 90, x = 0 and y = 0

Region represented by $\text{x}+\text{y}\leq50$.

The line x + y = 50 meets the coordinate axes at A(50, 0) and B(0, 50) respectively.

By joining these points we obtain the line 3x + 5y = 15.

Clearly (0, 0) satisfies the inequation $\text{x}+\text{y}\leq50$.

So, the region containing the origin represents the solution set of the inequation $\text{x}+\text{y}\leq50$.

Region represented by $3\text{x}+\text{y}\geq90$.

The line 3x + y = 90 meets the coordinate axes at C(30, 0) and D(0, 90) respectively.

By joining these points we obtain the line $3\text{x}+\text{y}\geq90$.

Clearly (0, 0) satisfies the inequation $3\text{x}+\text{y}\geq90$.

So, the region containing the origin represents the solution set of the inequation $3\text{x}+\text{y}\geq90$.

Region represented by $\text{x}\geq0$ and $\text{y}\geq0$:

Since, every point in the first quadrant satisfies these inequations.

So, the first quadrant is the region represented by the inequations $\text{x}\geq0$ and $\text{y}\geq0$:

The feasible region determined by the system of constraints,

$\text{x}+\text{y}\leq50,3\text{x}+\text{y}\geq90,\text{x}\geq0$ and $\text{y}\geq0$, are as follows.

The corner points of the feasible region are O(0, 0), C(30, 0), E(20, 30 ) and B(0, 50).

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

Therefore, the maximum value of Z is 1800 at the point (30, 0).

Hence, x = 30 and y = 0 is the optimal solution of the given LPP.

Thus, the optimal value of Z is 1800.

Converting the given inequations into equation,

-2x + y = 1, x = 2, x + y = 3, x= y = 0

Region represented by - 2x + y = 1:

Line - 2x + y = 1 meets coordinate axes at $\text{A}_1\Big(\frac{-1}{2},0\Big)$ and B₁(0, 1), clearly, (0, 0) satisfies $-2\text{x}+\text{y}\leq1$, so region containing origin represents $-2\text{x}+\text{y}\leq1$ in xy-plane.

Region represented by $\text{x}\leq2$:

Linex - 2 is parallel to y-axis and meets x-axis at A₃(2, 0).

Clearly, (0, 0) satisfies $\text{x}\leq2$, so region containing origin represents $\text{x}\leq2$ in xy-plane.

Region represented by $\text{x}+\text{y}\leq3$:

Line x + y - 3 meets coordinate axes at A₂(3, 0) and B₂(0, 3).

Clearly, (0, 0) satisfies $\text{x}+\text{y}\leq3$, so region containing origin represents $\text{x}+\text{y}\leq3$ in xy-plane.

Region represented $\text{x},\text{y}\geq0$:

It represents first quadrant in xy-plane.

So, shaded region OA₃PQ8, represents the feasible region.

Coordinates of P(2, 1) is obtained by solving x + y = 3 and x = 2, $\text{Q}\Big(\frac{2}{3},\frac{7}{3}\Big)$ by solving -2x + y = 1 and x + y = 3.

The value of Z = x + y at

$\text{O}(0, 0) = 0 + 0 = 0$

$\text{A}_3(2, 0) = 2 + 0 = 2$

$\text{P}(2, 1) = 2 +1 = 2$

$\text{Q}\Big(\frac{2}{3},\frac{7}{3}\Big)=\frac{2}{3}+\frac{7}{3}=3$

$\text{B}_1(0, 1) = 0 + 1 = 1$

So, maximum Z = 3 is at every point on the line joining PQ.

Hence, maximum z = 3 at x = 2 and y = 1 Or $\text{x}=\frac{2}{3}$ and $\text{y}=\frac{7}{3}.$

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(-2, 0) and B(0, 4).

Join these points to obtain the line -2x + y = 4.

Clearly, (0, 0) satisfies the inequation $-2\text{x}+\text{y}\leq4$.

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 $\text{x}+\text{y}\geq3$.

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 $\text{x}-2\text{y}\leq2$.

So, the region in xy-plane that contains the origin represents the solution set of the given equation.

Region represented by $\text{x}\geq0$ and $\text{y}\geq0$:

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 cornerpoint of the feasible region are B(0, 4), D(0, 3) and $\text{G}\Big(\frac{8}{3},\frac{1}{3}\Big)$.

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

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

Thus, the optimal value of Z is $\frac{29}{3}$.

$\text{As x}\geq0,\ \text{y}\geq0,$ therefore we shall shade the other inequalities in the first quadrant only.

Now $\text{x}+\text{y}\leq4$

Let x + y = 4

$\Rightarrow\frac{\text{x}}{\text{4}}+\frac{\text{y}}{4}=1$

Thus the line has 4 and 4 as intercepts along the axes. Now, (0, 0) satisfies the inequation, i.e., $0+0\leq4.$ Therefore, shaded region OAB is the feasible solution.

Its corners are O(0, 0), A(4, 0), B(0, 4)

Hence, max Z = 16 at x = 0, y = 4.

Converting the inequations into equations, ew obtain the lines.

2x + 4y = 8, 3x + y = 6, x + y =4, x = 0, y = 0.

These lines are drawn on a suitable scale and the feasible region of the LPP is shaded in the graph.

From the graph we can see the corner point as (0, 2) and (2, 0).

Now, solving the equations 3x + y = 6 and 2x +4y = 8 we get the values of x and y as $\text{x}=\frac{8}{5}$ and $\text{y}=\frac{6}{5}$.

Substituting $\text{x}=\frac{8}{5}$ and $\text{y}=\frac{6}{5}$ in Z = 2x + 5y we get,

$\text{z}=2\Big(\frac{8}{5}\Big)+5\Big(\frac{6}{5}\Big)$

$\text{z}=\frac{46}{5}$

Hence maximum value of Z is $\frac{46}{5}$ at $\text{x}=\frac{8}{5}$ and $\text{y}=\frac{6}{5}$.

Number of desktop model and portable model cannot be negative.

Therefore,

It is given that the monthly demand will not exist 250 units.

$\therefore$ x + y ≤ 250

Cost of desktop and portable model is Rs. 25,000 and Rs. 40,000 respectively.

Therefore, cost of x desktop model and y portable model is Rs. 25,000 and Rs. 40,000 respectively and he does not want to invest more than Rs. 70 lakhs.

25000x + 40000y ≤ 7000000

Profit on the desktop model is Rs. 4500 and on the portable model is Rs. 5000.

Therefore, profit made by x desktop model and y portable model is Rs. 4500x and Rs. 5000y respectively.

Total profit = Z = 4500x + 5000y

The mathematical form of the given LPP is:

Maximize Z = 4500x + 5000y

Subject to constraints:

x + y ≤ 250

25000x + 40000y ≤ 7000000

First we will convert inequations into equations as follows:

x + y = 250, 25000x + 40000y = 7000000, x = 0 and y = 0

Region represented by x + y ≤ 250:

The line x + y = 250 meets the coordinate axes at A(250, 0) and B(0, 250) respectively.

By joining these points we obtain the line x + y = 250.

Clearly (0, 0) satisfies the x + y = 250.

So, the region which contains the origin represents the solution set of the inequation x + y ≤ 250.

Region represented by 25000x + 40000y ≤ 7000000:

The line 25000x + 40000y = 7000000 meets the coordinate axes at C(280, 0) and D(0, 175) respectively.

By joining these points we obtain the line 25000x + 40000y = 7000000.

Clearly (0, 0) satisfies the inequation 25000x + 40000y ≤ 7000000.

So, the region which contains the origin represents the solution set of the inequation 25000x + 40000y ≤ 7000000.

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 ≤ 250, 25000x + 40000y ≤ 7000000, x ≥ 0 and y ≥ 0 are as follows.

The corner points are O(0, 0), D(0, 175), E(200, 50) and A(250, 0).

The values of the objective function Z at corner points of the feasible region are given in the following table:

Clearly, Z is maximum at x = 200 and y = 50 and the maximum value of Z at this point is 1150000.

Thus, 200 desktop models and 50 portable units should be sold to maximize the profit.

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 $\text{x}-\text{y}\geq0$ or $\text{x}\geq\text{y}$:

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 $\text{x}\geq\text{y}$.

So, it satisfy the given inequation.

Take a point (4, 5) to the left of the line x = y.

Here, $\text{x}\leq\text{y}$.

That means it does not satisfy the given inequation.

Region represented by $-\text{x}+2\text{y}\geq2$:

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 $-\text{x}+2\text{y}\geq2$.

So, the region in xy plane which does not contain the origin represents the solution set of the inequation $-\text{x}+2\text{y}\geq2$.

The line x = 3 is the line that passes through the point (3, 0) and is parallel to Y axis. $\text{x}\geq3$ 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. $\text{y}\geq4$ is the region below the line y = 4.

Region represented by $\text{x}\geq0$ and $\text{y}\geq0$:

Since, every point in the first quadrant satisfies these inequations.

So, the first quadrant is the region represented by the inequations $\text{x}\geq0$ and $\text{y}\geq0$.

The feasible region determined by the system of constraints $\text{x}-\text{y}\geq0,-\text{x}+2\text{y}\geq2,\text{x}\geq3,\text{y}\geq4,\text{x}\geq0,$and $\text{y}\geq0$ are as follows.

The corner points of the feasible region are $\text{C}\Big(3,\frac{5}{2}\Big),\text{D}(3, 3),\text{E}(4, 4)$ and $\text{F}(6, 4)$.

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.