Balbharati Maharashtra State Board12th Commerce Maths Solution Book PdfChapter 6 Linear Programming Ex 6.2 Questions and Answers.
Maharashtra State Board 12th Commerce Maths Solutions Chapter 6 Linear Programming Ex 6.2
Solve the following LPP by graphical method.
Question 1
Maharashtra Board Solution
Maximize z = 11x + 8y, Subject to x ≤ 4,y ≤ 6 x + y ≤ 6, x ≥ 0, y ≥ 0.
Solution & Step-by-Step Answer:
The feasible solution is AOBE Where A(4, 0) O(0, 0) B(0, 6) E is the point of intersection of x + y = 6 and x = 4. ∴ 4 + y = 6 ∴ y = 2 ∴ E = (4, 2) ∴ z is maximum at (4, 12) and the maximum value of z = 60
Question 2
Maharashtra Board Solution
Maximize z = 4x + 6y, Subject to 3x + 2y ≤ 12, x + y ≥ 4 x, y ≥ 0.
Solution & Step-by-Step Answer:
From figure, ABC is the feasible region Where A(0, 6) B(4, 0) C(0, 4) Maximum value of z = 36 at A(0, 6)
Question 3
Maharashtra Board Solution
Maximize z = 7x + 11y, Subject to 3x + 5y ≤ 26, 5x + 3y ≤ 30, x ≥ 0, y ≥ 0.
Solution & Step-by-Step Answer:
∴ AODE is the feasible region where A(0, 5.2) O(0, 0) D(6, 0) and E is the intersection of 3x + 5y = 26 and 5x + 3y = 30 For E, Solving 3x + 5y = 26 ……(i) 5x + 3y = 30 ……(ii) We get, x = 4.5, y = 2.5 ∴ E = (4.5, 2.5) ∴ Maximum value of z = 59 at E(4.5, 2.5)
Question 4
Maharashtra Board Solution
Maximize z = 10x + 25y, Subject to 0 ≤ x ≤ 3, 0≤ y ≤ 3, x + y ≤ 5.
Solution & Step-by-Step Answer:
The constraints can be written as, x ≤ 3, x ≥ 0, y ≥ 0, y ≤ 3, x + y ≤ 5 ABCDE is the feasible region where A(3, 0) B(0, 0) and C(0, 3) D is the intersection of y = 3 and x + 5y = 5 and E is the intersection of x = 3 and x + 7 = 5 For D, Solving y = 3 ………(i) x + y = 5 ……..(ii) We get x = 2, y = 3 ∴ D = (2, 3) For E, Solving x = 3 …….(i) x + y = 5 ……….(ii) We get x = 3, y = 2 ∴ E = (3, 2) ∴ Maximum value of z = 90 at D(2, 3)
Question 5
Maharashtra Board Solution
Maximize z = 3x + 5y, Subject to x + 4y ≤ 24, 3x + y ≤ 21, x + y ≤ 9, x ≥ 0, y ≥ 0.
Solution & Step-by-Step Answer:
OAGHD is the feasible region where O(0, 0), A(0, 6), D(7, 0) G is the intersecting point of x + 4y = 24 and x + y = 9 H is the intersecting points of 3x + y = 21 and x + y = 9. For G, Solving x + 4y = 24 …….(i) x + y = 9 ………(ii) We get, x = 4, y = 5 ∴ G (4, 5) For H, Solving x + y = 9 ………(i) 3x + y = 21 ……..(ii) We get x = 6, y = 3 ∴ H(6, 3) ∴ Maximum value of z = 37 at the point G(4, 5)
Question 6
Maharashtra Board Solution
Minimize z = 7x + y Subject to 5x + y ≥ 5, x + y ≥ 3, x ≥ 0, y ≥ 0
Solution & Step-by-Step Answer:
AED is the feasible region where A(0, 5) D(3, 0) and E is the point of intersection of 5x + y = 5 and x + y = 3. For E, Solving 5x + y = 5 ………(i) x + y = 3 ……..(ii) We get, x = , y = ∴ E(, ) ∴ Minimum value of z = 5 at A(0, 5)
Question 7
Maharashtra Board Solution
Minimize z = 8x + 10y, Subject to 2x + y ≥ 7, 2x + 3y ≥ 15,y ≥ 2, x ≥ 0, y ≥ 0
Solution & Step-by-Step Answer:
AEG is the feasible solution where A(0, 7) F is the point of intersection of 2x + y = 7 and 2x + 3y = 15 G is the point of intersection of y = 2 and 2x + 3y = 15 For F, Solving 2x + y = 7 ……..(i) 2x + 3y = 15 ……..(ii) We get x = , y = 4 ∴ F = (, 4) For G, Solving 2x + 3y = 15 ……..(i) y = 2 …….(ii) We get x = 4.5, y = 2 ∴ G = (4.5, 2) ∴ Minimum value of z = 52 at F(, 4)
Question 8
Maharashtra Board Solution
Minimize z = 6x + 2y, Subject to x + 2y ≥ 3, x + 4y ≥ 4, 3x + y ≥ 3, x ≥ 0, y ≥ 0.
Solution & Step-by-Step Answer:
DGHB is the feasible region where D(0, 3), B(4, 0) G is the point of intersection of 3x + y = 3 and x + 2y = 3 and H is the point of intersection of x + 2y = 3 and x + 4y = 4 For G, Solving 3x + y = 3 ……(i) x + 2y = 3 ………(ii) W e get x = , y = ∴ G(, ) For H, Solving x + 2y = 3 ……..(i) x + 4y = 4 ………(ii) We get, x = 2, y = ∴ H(2, ) ∴ Minimum value of z = 22.5 at H(2, )
