Maths Solve the Following Question.(5 Marks)

Let the firm manufactures x units of item A and y units of item B. Firm can make profit of ₹ 20 per unit of A and ₹ 30 per unit of B. Hence, the total profit is z = ₹ (20x + 30y). This is the objective function which is to be maximized. The constraints are as per the following table :

From the table, the constraints are 3x + 2y ≤ 210, 2x + 4y ≤ 300 Since, number of items cannot be negative, x ≥ 0, y ≥ 0. Hence, the mathematical formulation of given LPP is : Maximize z = 20x + 30y, subject to 3x + 2y ≤ 210, 2x + 4y ≤ 300, x ≥ 0, y ≥ 0. We draw the lines AB and CD whose equations are 3x + 2y = 210 and 2x + 4y = 300 respectively.

The feasible region is OAPDO which is shaded in the graph. The vertices of the feasible region are O (0, 0), A (70, 0), P and D (0, 75). P is the point of intersection of the lines 2x + 4y = 300 … (1) and 3x + 2y = 210 … (2) Multiplying equation (2) by 2, we get 6x + 4y = 420 Subtracting equation (1) from this equation, we get ∴ 4x = 120 ∴ x = 30 Substituting x = 30 in (1), we get 2(30) + 4y = 300 ∴ 4y = 240 ∴ y = 60 ∴ P is (30, 60) The values of the objective function z = 20x + 30y at these vertices are z(O) = 20(0) + 30(0) = 0 + 0 = 0 z(A) = 20(70) + 30(0) = 1400 + 0 = 1400 z(P) = 20(30) + 30(60) = 600 + 1800 = 2400 z(D) = 20(0) + 30(75) = 0 + 2250 = 2250 ∴ z has the maximum value 2400 when x = 30 and y = 60. Hence, the firm should manufactured 30 units of item A and 60 units of item B to get the maximum profit of ₹ 2400.

Let the firm manufactures x units of product A and y units of product B. The profit earned per unit of A is ₹3 and B is ₹ 4. Hence, the total profit is z = ₹ (3x + 4y). This is the linear function which is to be maximized. Hence, it is the objective function. The constraints are as per the following table :

From the table, the constraints are x + y ≤ 450, 2x + y ≤ 600 Since, the number of gift items cannot be negative, x ≥ 0, y ≥ o. ∴ the mathematical formulation of LPP is, Maximize z = 3x + 4y, subject to x + y ≤ 450, 2x + y ≤ 600, x ≥ 0, y ≥ 0. Now, we draw the lines AB and CD whose equations are x + y = 450 and 2x + y — 600 respectively.

The feasible region is OCPBO which is shaded in the graph. The vertices of the feasible region are O(0, 0), C(300, 0), P and B (0, 450). P is the point of intersection of the lines 2x + y = 600 … (1) and x + y = 450 … (2) On subtracting, we get ∴ x = 150 Substituting x = 150 in equation (2), we get 150 + y = 450 ∴ y = 300 ∴ P = (150, 300) The values of the objective function z = 3x + 4y at these vertices are z(O) = 3(0) + 4(0) = 0 + 0 = 0 z(C) = 3(300) + 4(0) = 900 + 0 = 900 z(P) = 3(150) + 4(300) = 450 + 1200 = 1650 z(B) = 3(0) + 4(450) = 0 + 1800 = 1800 ∴ z has the maximum value 1800 when x = 0 and y = 450 Hence, the firm gets maximum profit of ₹ 1800 if it manufactures 450 units of product B and no unit product A.

Let x: number of gift item A y: number of gift item B As numbers of the items are never negative x ≥ 0; y ≥ 0

Total time required for the cutter = 4x + 2y Maximum available time 208 hours ∴ 4x+ 2y ≤ 208 Total time required for the finisher 2x +4y Maximum available time 152 hours 2x + 4y ≤ 152 Total Profit is 75x + 125y ∴ L.P.P. of the above problem is Minimize z = 75x + 125y Subject to 4x+ 2y ≤ 208 2x + 4y ≤ 152 x ≥ 0; y ≥ 0 Graphical solution

Corner points Now, Z at x = (75x + 125y) O(0, 0) = 75 × 0 + 125 × 0 = 0 A(52,0) = 75 × 52 + 125 × 0 = 3900 B(44, 16) = 75 × 44 + 125 × 16 = 5300 C(0, 38) = 75 × 0 + 125 × 38 = 4750 A person should make 44 items of type A and 16 Uems of type Band his returns are ₹ 5,300.

Let the company buy x units of compound I and y units of compound II. Then the total cost is z = ₹(800x + 640y). This is the objective function which is to be minimized. The constraints are as per the following table :

From the table, the constraints are 4x + 2y ≥ 16, 12x + 2y ≥ 24, 2x + 6y ≥ 18. Also, the number of units of compound I and compound II cannot be negative. ∴ x ≥ 0, y ≥ 0. ∴ the mathematical formulation of given LPP is Minimize z = 800x + 640y, subject to 4x + 2y ≥ 16, 12x + 2y ≥ 24, 2x + 6y ≥ 18, x ≥ 0, y ≥ 0. First we draw the lines AB, CD and EF whose equations are 4x + 2y = 16, 12x + 2y = 24 and 2x + 6y = 18

The feasible region is shaded in the graph. The vertices of the feasible region are E(9, 0), P, Q, and D(0, 12). P is the point of intersection of the lines 2x + 6y = 18 … (1) and 4x + 2y = 16 … (2) Multiplying equation (1) by 2, we get 4x + 12y = 36 Subtracting equation (2) from this equation, we get 10y = 20 ∴ y = 2 ∴ from (1), 2x + 6(2) = 18 ∴ 2x = 6 ∴ x = 3 ∴ P = (3, 2) Q is the point of intersection of the lines 12x + 2y = 24 … (3) and 4x + 2y = 16 … (2) On subtracting, we get 8x = 8 ∴ x = 1 ∴ from (2), 4(1) + 2y = 16 ∴ 2y = 12 ∴ y = 6 ∴ Q = (1, 6) The values of the objective function z = 800x + 640y at these vertices are z(E) = 800(9)+ 640(0) =7200 + 0 = 7200 z(P) = 800(3) + 640(2) = 2400 + 1280 = 3680 z(Q) = 800(1) + 640(6) =800 + 3840 =4640 z(D) = 800(0) + 640(12) = 0 + 7680 = 7680 ∴ the minimum value of z is 3680 at the point (3, 2). Hence, the company should buy 3 units of compound I and 2 units of compound II to have the minimum cost of ₹ 3680.

Let the factory produce x units of chemical A and y units of chemical B. Then the total cost is z = ₹ (4x + 6y). This is the objective function which is to be minimized. From the given table, the constraints are x + 2y ≥ 80, 3x + y ≥ 75. Also, the number of units x and y of chemicals A and B cannot be negative. ∴ x ≥ 0, y ≥ 0. ∴ the mathematical formulation of given LPP is Minimize z = 4x + 6y, subject to x + 2y ≥ 80, 3x + y ≥ 75, x ≥ 0, y ≥ 0. First we draw the lines AB and CD whose equations are x + 2y = 80 and 3x + y = 75 respectively.

The feasible region is shaded in the graph. The vertices of the feasible region are A (80, 0), P and D (0, 75). P is the point of intersection of the lines x + 2y = 80 … (1) and 3x + y = 75 … (2) Multiplying equation (2) by 2, we get 6x + 2 y = 150 Subtracting equation (1) from this equation, we get 5x = 70 ∴ x = 14 ∴ from (2), 3(14) + y = 75 ∴ 42 + y = 75 ∴ y = 33 ∴ P = (14, 33) The values of the objective function z = 4x + 6y at these vertices are z(a) = 4(80)+ 6(0) =320 + 0 = 320 z(P) = 4(14)+ 6(33) = 56+ 198 = 254 z(D) = 4(0) + 6(75) = 0 + 450 = 450 ∴ the minimum value of z is 254 at the point (14, 33). Hence, 14 units of chemical A and 33 units of chemical B are to be produced in order to have the j minimum cost of ₹ 254.

Let x bicycles and y tricycles are to be manu¬factured. Then the total profit is z = ₹ (180x + 220y) This is a linear function which is to be maximized. Hence, it is the objective function. The constraints are as per the following table :

From the table, the constraints are 6x + 4y ≤ 120, 3x +10y ≤ 180 Also, the number of bicycles and tricycles cannot be i negative. ∴ x ≥ 0, y ≥ 0. Hence, the mathematical formulation of given LPP is : Maximize z = 180x + 220y, subject to 6x + 4y ≤ 120, 3x + 10y ≤ 180, x ≥ 0, y ≥ 0. First we draw the lines AB and CD whose equations are 6x + 4y = 120 and 3x + 10y = 180 respectively.

The feasible region is OAPDO which is shaded in the graph. The vertices of the feasible region are O(0, 0), A(20, 0) P and D(0, 18). P is the point of intersection of the lines 3x + 10y = 180 … (1) and 6x + 4y = 120 … (2) Multiplying equation (1) by 2, we get 6x + 20y = 360 Subtracting equation (2) from this equation, we get 16y = 240 ∴ y = 15 ∴ from (1), 3x + 10(15) = 180 ∴ 3x = 30 ∴ x = 10 ∴ P = (10, 15) The values of the objective function z = 180x + 220y at these vertices are z(O) = 180(0) + 220(0) = 0 + 0 = 0 z(a) = 180(20) + 220(0) = 3600 + 0 = 3600 z(P) = 180(10) + 220(15) = 1800 + 3300 = 5100 z(D) = 180(0) +220(18) = 3960 ∴ the maximum value of z is 5100 at the point (10, 15). Hence, 10 bicycles and 15 tricycles should be manufactured in order to have the maximum profit of ₹ 5100.

We first draw the lines AB, CD and EF whose equations are 3x + y = 27, x + y = 21, x + 2y = 30 respectively.

The feasible region is XEPQBY which is shaded in the graph. The vertices of the feasible region are E (30,0), P, Q and B (0,27). P is the point of intersection of the lines x + 2y = 30 … (1) and x + y = 21 … (2) On subtracting, we get y = 9 Substituting y = 9 in (2), we get x + 9 = 21 ∴ x = 12 ∴ P is (12, 9) Q is the point of intersection of the lines x + y = 21 … (2) and 3x + y = 27 … (3) On subtracting, we get 2x = 6 ∴ x = 3 Substituting x = 3 in (2), we get 3 + y = 21 ∴ y = 18 ∴ Q is (3, 18). The values of the objective function z = 4x + 2y at these vertices are z(E) = 4(30) + 2(0) = 120 + 0 = 120 z(P) = 4(12) + 2(9) = 48 + 18 = 66 z(Q) = 4(3) + 2(18) = 12 + 36 = 48 z(B) = 4(0) + 2(27) = 0 + 54 = 54 ∴ z has minimum value 48, when x = 3 and y = 18.

We first draw the lines AB and CD whose equations are x + 2y = 40 and 3x + 2y = 60 respectively.

The feasible region is OCPBO which is shaded in the graph. The vertices of the feasible region are O (0, 0), C (20, 0), P and B (0, 20). P is the point of intersection of the lines. 3x + 2y = 60 … (1) and x + 2y = 40 … (2) On subtracting, we get 2x = 20 ∴ x = 10 Substituting x = 10 in (2), we get 10 + 2y = 40 ∴ 2y = 30 ∴ y = 15 ∴ P is (10, 15) The values of the objective function z = 60x + 50y at these vertices are z(O) = 60(0) + 50(0) = 0 + 0 = 0 z(C) = 60(20) + 50(0) = 1200 + 0 = 1200 z(P) = 60(10) + 50(15) = 600 + 750 = 1350 z(B) = 60(0) + 50(20) = 0 + 1000 = 1000 . ∴ z has maximum value 1350 at x = 10, y = 15.

First we draw the lines AB whose equation is x – y = 3.

The feasible region is shaded which is unbounded. Therefore, the value of objective function can be in- j creased indefinitely. Hence, this LPP has unbounded solution.

First we draw the lines AB and CD whose equations are 3x + y = 27 and x + y = 21 respectively.

The feasible region is OAPDO which is shaded region in the graph. The vertices of the feasible region are 0(0, 0), A (9, 0), P and D(0, 21). P is the point of intersection of lines 3x + y = 27 … (1) and x + y = 21 … (2) On substracting, we get 2x = 6 ∴ x = 3 Substituting x = 3 in equation (1), we get 9 + y = 27 ∴ y = 18 ∴ P = (3, 18) The values of the objective function z = 4x + 2y at these vertices are z(O) = 4(0) + 2(0) = 0 + 0 = 0 z(a) = 4(9) + 2(0) = 36 + 0 = 36 z(P) = 4(3) + 2(18) = 12 + 36 = 48 z (D) = 4(0) + 2(21) = 0 + 42 = 42 ∴ 2 has minimum value 48 when x = 3, y = 18.

First we draw the lines AB, CD and EF whose equations are 2x + y = 7, 2x + 3y = 15 and y = 2 respectively.

The feasible region is EPQBY which is shaded in the graph. The vertices of the feasible region are P, Q and B(0,7). P is the point of intersection of the lines 2x + 3y = 15 and y = 2. Substituting y – 2 in 2x + 3y = 15, we get 2x + 3(2) = 15 ∴ 2x = 9 ∴ x = 4.5 ∴ P = (4.5, 2) Q is the point of intersection of the lines 2x + 3y = 15 … (1) and 2x + y = 7 … (2) On subtracting, we get 2y = 8 ∴ y = 4 ∴ from (2), 2x + 4 = 7 ∴ 2x = 3 ∴ x = 1.5 ∴ Q = (1.5, 4) The values of the objective function z = 8x + 10y at these vertices are z(P) = 8(4.5) + 10(2) = 36 + 20 = 56 z(Q) = 8(1.5) + 10(4) = 12 + 40 = 52 z(B) = 8(0) +10(7) = 70 ∴ z has minimum value 52, when x = 1.5 and y = 4

First we draw the lines AB, CD and EF whose equations are x + 4y = 24, 3x + y = 21 and x + y = 9 respectively.

The feasible region is OCPQBO which is shaded in the graph. The vertices of the feasible region are O (0, 0), C (7, 0), P, Q and B (0, 6). P is the point of intersection of the lines 3x + y = 21 … (1) and x + y = 9 … (2) On subtracting, we get 2x = 12 ∴ x = 6 Substituting x = 6 in equation (2), we get 6 + y = 9 ∴ y = 3 ∴ P = (6, 3) Q is the point of intersection of the lines x + 4y = 24 … (3) and x + y = 9 … (2) On subtracting, we get 3y = 15 ∴ y = 5 Substituting y = 5 in equation (2), we get x + 5= 9 ∴ x = 4 ∴ Q = (4, 5) ∴ the corner points of the feasible region are 0(0,0), C(7, 0), P (6, 3), Q (4, 5) and B (0, 6). The values of the objective function 2 = 3x + 5y at these corner points are z(O) = 3(0)+ 5(0) = 0 + 0 = 0 z(C) = 3(7) + 5(0) = 21 + 0 = 21 z(P) = 3(6) + 5(3) = 18 + 15 = 33 z(Q) = 3(4) + 5(5) = 12 + 25 = 37 z(B) = 3(0)+ 5(6) = 0 + 30 = 30 ∴ z has maximum value 37, when x = 4 and y = 5.

First we draw the lines AB, CD and EF whose equations are x = 3, y = 3 and x + y = 5 respectively.

The feasible region is OAPQDO which is shaded in the i graph. The vertices of the feasible region are O (0, 0), A (3, 0), P, Q and D(0, 3). t P is the point of intersection of the lines x + y = 5 and x = 3. Substituting x = 3 in x + y = 5, we get 3 + y = 5 ∴ y = 2 ∴ P is (3, 2) Q is the point of intersection of the lines x + y = 5 and y = 3 Substituting y = 3 in x + y = 5, we get x + 3 = 5 ∴ x = 2 ∴ Q is (2, 3) The values of the objective function z = 10x + 25y at these vertices are z(O) = 10(0) + 25(0) = 0 + 0 = 0 z(a) = 10(3) + 25(0) = 30 + 0 = 30 z(P) = 10(3) + 25(2) = 30 + 50 = 80 z(Q) = 10(2) + 25(3) = 20 + 75 = 95 z(D) = 10(0)+ 25(3) = 0 + 75 = 75 ∴ z has maximum value 95, when x = 2 and y = 3.

First we draw the lines AB and AC whose equations are 3x + 2y = 12 and x + y = 4 respectively.

The feasible region is the ∆ABC which is shaded in the graph. The vertices of the feasible region (i.e. corner points) are A (4, 0), B (0, 6) and C (0, 4). The values of the objective function z = 4x + 6y at these vertices are z(a) = 4(4) + 6(0) = 16 + 0 = 16 z(B) = 4(0)+ 6(6) = 0 + 36 = 36 z(C) = 4(0) + 6(4) = 0 + 24 = 24 ∴ has maximum value 36, when x = 0, y = 6.

First we draw the lines AB, CD and ED whose equations are x = 4, y = 6 and x + y = 6 respectively.

The feasible region is shaded portion OAPDO in the graph. The vertices of the feasible region are O (0, 0), A (4, 0), P and D (0, 6) P is point of intersection of lines x + y = 6 and x = 4. Substituting x = 4 in x + y = 6, we get 4 + y = 6 ∴ y = 2 ∴ P is (4, 2). ∴ the corner points of feasible region are O (0, 0), A (4, 0), P(4, 2) and D(0 ,6). The values of the objective function z = 11x + 8y at these vertices are z (O) = 11(0) + 8(0) = 0 + 0 = 0 z(a) = 11(4) + 8(0) = 44 + 0 = 44 z (P) = 11(4) + 8(2) = 44 + 16 = 60 z (D) = 11(0) + 8(2) = 0 + 16 = 16 ∴ z has maximum value 60, when x = 4 and y = 2.

Let the diet of sick person include x units of food A and y units of food B. Then x ≥ 0, y ≥ 0. The prices of food A and B are ₹ 4.5 and ₹ 3.5 per unit respectively. Therefore, the total cost is z = ₹ (4.5x + 3.5y) This is the linear function which is to be minimized. Hence, it is objective function. The constraints are as per the following table :

From the table, the sick person’s diet will include (4x + 6y) units of fats, (14x + 12y) units of carbohydrates and (8x + 8y) units of proteins. The minimum requirements of these ingredients are 18 units, 28 units and 14 units respectively. Therefore, the constraints are 4x + 6y ≥ 18, 14x + 12y ≥ 28, 8x + 8y ≥ 14. Hence, the given LPP can be formulated as Minimize z = 4.5x + 3.5y, subject to 4x + 6y ≥ 18, 14x + 12y ≥ 28, 8x + 8y ≥ 14, x ≥ 0, y ≥ 0.

Let the company manufactures x units of fertilizers F1 and y units of fertilizers F1. Then the total profit to the company is z = ₹(500x + 750y). This is a linear function that is to be maximized. Hence, it is an objective function.

The raw material A required for x units of Fertilizers F1 and y units of Fertilizers F2 is 2x + Since the maximum availability of A is 40, we have the first constraint as 2x + 3y ≤ 40. Similarly, considering the raw material B, we have x + 4y ≤ 70. Since, x and y cannot be negative, we have, x ≥ 0, y ≥ 0. Hence, the given LPP can be formulated as: Maximize z = 500x + 750y, subject to 2x + 3y ≤ 40, x + 4y ≤ 70, x ≥ 0, y ≥ 0.

Let the number of packages of bulbs produced by manufacturer be x and packages of tubes be y. The manufacturer earns a profit of ₹ 13.5 per package of bulbs and ₹ 55 per package of tubes. Therefore, his total profit is p = ₹ (13.5x + 55y) This is a linear function which is to be maximized. Hence, it is the objective function. The constraints are as per the following table :

From the table, the total time required for Machine M1 is (x + 2y) hours and for Machine M2 is (3x + 4y) hours. Given Machine M1 and M2 are available for atmost 10 hours and 12 hours a day respectively. Therefore, the constraints are x + 2y ≤ 10, 3x + 4y ≤ 12. Since, x and y cannot be negative, we have, x ≥ 0, y ≥ 0. Hence, the given LPP can be formulated as : Maximize p = 13.5x + 55y, subject to x + 2y ≤ 10, 3x + 4y ≤ 12, x ≥ 0, y ≥ 0.

Let the company prints x magazine of type A and y magazine of type B. Profit on sale of magazine A is ₹ 10 per copy and magazine B is ₹ 15 per copy. Therefore, the total earning z of the company is z = ₹ (10x + 15y). This is a linear function which is to be maximized. Hence, it is the objective function. The constraints are as per the following table:

From the table, the total time required for Machine I is (2x + 3y) hours, for Machine II is (5x + 2y) hours and for Machine III is (2x + 6y) hours. The machines I, II, III are available for 36,50 and 60 hours per week. Therefore, the constraints are 2x + 3y ≤ 36, 5x + 2y ≤ 50, 2x + 6y ≤ 60. Since x and y cannot be negative. We have, x ≥ 0, y ≥ 0. Hence, the given LPP can be formulated as : Maximize z = 10x + 15y, subject to 2x + 3y ≤ 36, 5x + 2y ≤ 50, 2x + 6y ≤ 60, x ≥ 0, y ≥ 0.

Let the company manufactures x units of chemical A and y units of chemical B. Then the total profit f to the company is p = ₹ (350x + 400y). This is a linear function which is to be maximized. Hence, it is the objective function. The constraints are as per the following table:

The raw material P required for x units of chemical A and y units of chemical B is 3x + 2y. Since, the maximum availability of P is 120, we have the first constraint as 3x + 2y ≤ 120. Similarly, considering the raw material Q, we have : 2x + 5y ≤ 160. Since, x and y cannot be negative, we have, x ≥ 0, y ≥ 0. Hence, the given LPP can be formulated as : Maximize p = 350x + 400y, subject to 3x + 2y ≤ 120, 2x + 5y ≤ 160, x ≥ 0, y ≥ 0.

Let x units of fodder 1 and y units of fodder 2 be prescribed. The cost of fodder 1 is ₹ 3 per unit and cost of fodder 2 is ₹ 2 per unit. ∴ total cost is z = 3x + 2y This is the linear function which is to be minimized. Hence it is the objective function. The constraints are as per the following table :

From table fodder contains (2x + y) units of nutrients A, (2x + 3y) units of nutrients B and (x + y) units of nutrients C. The minimum requirements of these nutrients are 14 units, 22 units and 1 unit respectively. Therefore, the constraints are 2x + y ≥ 14, 2x + 3y ≥ 22, x + y ≥ 1 Since, number of units (i.e. x and y) cannot be negative, we have, x ≥ 0, y ≥ 0. Hence, the given LPP can be formulated as Minimize z = 3x + 2y, subject to 2x + y ≥ 14, 2x + 3y ≥ 22, x + y ≥ 1, x ≥ 0, y ≥ 0.

Let the number of gadgets A produced by the firm be x and the number of gadgets B produced by the firm be y. The profit on the sale of A is ₹ 30 per unit and on the sale of B is ₹ 20 per unit. ∴ total profit is z = 30x + 20y. This is a linear function which is to be maximized. Hence it is the objective function. The constraints are as per the following table :

From the table total man hours of labour required for x units of gadget A and y units of gadget B in foundry is (10x + 6y) hours and total man hours of labour required in machine shop is (5x + 4y) hours. Since, maximum time avilable in foundry and machine shops are 60 hours and 35 hours respectively. Therefore, the constraints are 10x + 6y ≤ 60, 5x + 4y ≤ 35. Since, x and y cannot be negative, we have x ≥ 0, y ≥ 0. Hence, the given LPP can be formulated as : Maximize z = 30x + 20y, subject to 10x + 6y ≤ 60, 5x + 4y ≤ 35, x ≥ 0, y ≥ 0.

Let x be the number of tables and y be the number of chairs. Then x ≥ 0, y ≥ 0. The dealer has a space to store at most 60 pieces. ∴ x + y ≤ 60 Since, the cost of each table is ₹ 150 and that of each chair is ₹ 750, the total cost of x tables and y chairs is 150x + 750y. Since the dealer has ₹ 15,000 to invest, 150x + 750y ≤ 15,000 Hence the system of inequations are x + y ≤ 60, 150x + 750y ≤ 15000, x ≥ 0, y ≥ 0. First we draw the lines AB and CD whose equations are x + y = 60 and 150x + 750y = 15,000 respectively.

The feasible solution is OAPDO. which is shaded in the graph.

Let the company produces x units of article A and y units of article B. The given data can be tabulated as:

Inequations are : x + 2y ≤ 4 and 3x + 2y ≤ 6 x and y are number of items, x ≥ 0, y ≥ 0 First we draw the lines AB and CD whose equations are x + 2y = 4 and 3x + 2y = 6 respectively.

The feasible solution is OCPBO. which is shaded in the graph.

First we draw the lines AB, CD and EF whose equations are x – 2y = 2, x + y = 3 and -2x + y = 4 respectively.

The feasible solution is shaded in the graph.

First we draw the lines AB, CD, EF and GH whose equations are x + y = 5, 2x + y = 4, x = 3 and y = 3 respectively.

The feasible solution is CEPQRC. which is shaded in the graph.

First we draw the lines AB, CD and EF whose equations are x + 4y = 24, 3x + y = 21 and x + y = 9 respectively.

The feasible solution is OCPQBO. which is shaded in the graph.

First we draw the lines AB, CD and EF whose equations are 3x + 4 y = 12, 4x + 7y = 28 and y = 1 respectively.

The feasible solution is PQDBP. which is shaded in the graph.

First we draw the lines AB and CB whose equations are 2x + 3y = 6 and x + y = 2 respectively.

The feasible solution is ∆ABC which is shaded in the graph.

First we draw the lines AB and CD whose equations are 3x + 2y = 18 and 2x + y = 10 respectively

The feasible solution is OCPBO which is shaded in the graph.

First we draw the lines AB and CD whose equations are 2x + y = 5 and x – y = 1 respectively.

The solution set of the given system of inequations is shaded in the graph. CategoriesClass 12

First we draw the lines AB and CD whose equations are 2x + 3y = 6 and x + 4y = 4 respectively.

The solution set of the given system of inequalities is shaded in the graph.

First we draw the lines AB and CD whose equations are x + y = 6 and x + 2y = 10 respectively.

The solution set of the given system of inequalities is shaded in the graph.

First we draw the lines AB and CD whose equations are x – y = 2 and x + 2y = 8 respectively.

The solution set of the given system of inequalities is shaded in the graph.

First we draw the lines AB and AC whose equations are 2x + y = 2 and x – y = 1 respectively

The solution set of the given system of inequalities is shaded in the graph.

Consider the line whose equation is 5x – 3y = 0. The constant term is zero, therefore this line is passing through the origin. ∴ one point on the line is the origin O = (0, 0). To find the other point, we can give any value of x and get the corresponding value of y. Put x = 3, we get 15 – 3y = 0, i.e. y = 5 ∴ A ≡ (3, 5) is another point on the line. Draw the line OA. To find the solution set, we cannot check 0(0, 0), as it is already on the line. We can check any other point which is not on the line. Let us check the point (1, -1). When x = 1, y = -1 then 5x – 3y = 5 + 3 = 8 which is neither less nor equal to zero. ∴ 5x – 3y ≰ 0 in this case. Hence (1, -1) will not lie in the required region. Therefore, the required region is the upper side which is shaded in the graph.

This is the solution set of 5x – 3y ≤ 0.

Consider the line whose equation is 3x + 2y = 0. The constant term is zero, therefore this line is passing through the origin. ∴ one point on the line is O ≡ (0, 0). To find the another point, we can give any value of x and get the corresponding value of y. Put x = 2, we get 6 + 2y = 0 i.e. y = – 3 ∴ A = (2, -3) is another point on the line. Draw the line OA. To find the solution set, we cannot check (0, 0) as it is already on the line. We can check any other point which is not on the line. Let us check the point (1, 1)

When x = 1, y = 1, then 3x + 2y = 3 + 2 = 5 which is greater than zero. ∴ 3x + 2y > 0 in this case. Hence (1, 1) lies in the required region. Therefore, the required region is the upper side which is shaded in the graph. This is the solution set of 3x + 2y ≥ 0.

Consider the line whose equation is 2x – 5y = 10. To find the points of intersection of this line with the coordinate axes. Put y = 0, we get 2x = 10, i.e. x = 5. ∴ A = (5, 0) is a point on the line. Put x = 0, we get -5y = 10, i.e. y = -2 ∴ B = (0, -2) is another point on the line.

Draw the line AB joining these points. This line J divide the plane in two parts. 1. Origin side 2. Non-origin side To find the solution set, we have to check the position of the origin (0, 0) with respect to the line. When x = 0, y = 0, then 2x – 5y = 0 which is neither greater nor equal to 10. ∴ 2x – 5y ≱ 10 in this case. Hence (0, 0) will not lie in the required region. Therefore, the given inequality is the non-origin side, which is shaded in the graph. This is the solution set of 2x – 5y ≥ 10.

Consider the line whose equation is x + 2y = 6. To find the points of intersection of this line with the coordinate axes. Put y = 0, we get x = 6. ∴ A = (6, 0) is a point on the line. Put x = 0, we get 2y = 6, i.e. y = 3 ∴ B = (0, 3) is another point on the line.

Draw the line AB joining these points. This line divide the line into two parts. 1. Origin side 2. Non-origin side To find the solution set, we have to check the position of the origin (0, 0) with respect to the line. When x = 0, y = 0, then x + 2y = 0 which is less than 6. ∴ x + 2y ≤ 6 in this case. Hence, origin lies in the required region. Therefore, the given inequality is the origin side which is shaded in the graph. This is the solution set of x + 2y ≤ 6.