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## Example 2 - Chapter 12 Class 12 Linear Programming

Last updated at May 29, 2023 by Teachoo Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class

Example 2 Solve the following linear programming problem graphically: Minimise Z = 200 x + 500 y subject to the constraints: x + 2y ≥ 10 3x + 4y ≤ 24 x ≥ 0, y ≥ 0 Minimize Z = 200x + 500y Subject to, x + 2y ≥ 10 3x + 4y ≤ 24 x ≥ 0, y ≥ 0 Hence Z is minimum at (4, 3) ## Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.

## Solve the following linear programming problem graphically : Minimise Z = 200 x + 500 y . . . ( 1 ) subject to the constraints: x + 2 y ≥ 10 . . . ( 2 ) 3 x + 4 y ≤ 24 . . . ( 3 ) x ≥ 0 , y ≥ 0 . . . ( 4 ) .

The shaded region in fig is the feasible region a b c determined by the system of constraints ( 2 ) to ( 4 ) , which is bounded. the coordinates of corner points corner point corresponding value of z ( 0 , 5 ) 2500 ( 4 , 3 ) 2300 ← m i n i m u m ( 0 , 6 ) 3000 a , b and c are ( 0 , 5 ) , ( 4 , 3 ) and ( 0 , 6 ) respectively. now we evaluate z = 200 x + 500 y at these points. hence, minimum value of z is 2300 attained at the point ( 4 , 3 ) ..

Solve the following linear programming problem graphically. Minimize Z = 3x+5y Subject to the constraints x + 2 y ≥ 10 , x + y ≥ 6 , 3 x + y ≥ 8 , x ≥ 0 , y ≥ 0. T o m i n i m i z e : z = 3 x + 5 y S u b j e c t t o t h e c o n s t r a i n t s x + 2 y ≥ 10 , x + y ≥ 6 , 3 x + y ≥ 8 , x ≥ 0 , y ≥ 0

Solve the following linear programming problem graphically: Maximise Z = 7 x + 10 y subject to the constraints 4 x + 6 y ≤ 240 6 x + 3 y ≤ 240 x ≥ 10 x ≥ 0 , y ≥ 0

Solve the following Linear Programming Problems graphically: Minimise Z = − 3 x + 4 y subject to x + 2 y ≤ 8 , 3 x + 2 y ≤ 12 , x ≥ 0 , y ≥ 0 .

Solve the following linear programming problem graphically: Maximize: Z = 60 x + 40 y subject to the constraints:

Solve the following Linear Programming problems graphically: 1. Maximize Z = 3 x + 4 y Subject to the constraints : x + y ≤ 4 , x ≥ 0 , y ≥ 0 2. Minimize Z = − 3 x + 4 y subject to x + 2 y ≤ 8 , 3 x + 2 y ≤ 12 , x ≥ 0 , y ≥ 0  ## Solve the following linear programming problem graphically: Minimize Z = 200 x + 500 y Subject to the constraints: x + 2 y ≥ 10 3 x + 4 y ≤ 24 x ≥ 0 , y ≥ 0

Minimise z = 200 x + 500 y subject to x + 2 y ≥ 10 , 3 x + 4 y ≤ 24 , x ≥ 0 , y ≥ 0 x + 2 y = 10 x 0 10 y 5 0 3 x + 4 y = 24 x 0 8 y 6 0 corner points value of z = 200 x + 500 y ( 0 , 6 ) 3000 ( 0 , 5 ) 2500 ( 4 , 3 ) 2300 ∴ minimum value of z is 2300 at ( 4 , 3 ) .. ## Solve the following linear programming problem graphically: Minimise Z = 200x + 500y subject to the constraints  x + 2y ≥ 10,  3x + 4 y ≤ 24,   x ≥ 0, y ≥ 0 ∴  CML is the feasible region, which is bounded. The comer points are C(4, 3), M(0, 6), L(0, 5). At C(4, 3), Z = 800 + 1500 = 2300 At M(0, 6), Z = 0 + 3000 = 3000 At L(0, 5), Z = 0 + 2500 = 2500 ∴ minimum value = 2300 at (4, 3).

## Linear Programming

Mathematics part ii.

• Suggest Edit ## Solve the following linear programming problem graphically. Minimize Z = 200x1 + 500x2 subject to the constraints: x1 + 2x2 ≥ 10; 3x1 + 4x2 ≤ 24 and x1 ≥ 0, x2 ≥ 0. - Business Mathematics and Statistics

Solve the following linear programming problem graphically.

Minimize Z = 200x 1 + 500x 2 subject to the constraints: x 1 + 2x 2 ≥ 10; 3x 1 + 4x 2 ≤ 24 and x 1 ≥ 0, x 2 ≥ 0.

## Solution Show Solution

Since the decision variables, x 1 and x 2 are non-negative, the solution lies in the I quadrant of the plane.

Consider the equations

x 1 + 2x 2 = 10

3x 1 + 4x 2 = 24 The feasible region is ABC and its co-ordinates are A(0, 5) C(0, 6) and B is the point of intersection of the lines

x 1 + 2x 2 = 10 ..........(1)

3x 1 + 4x 2 = 24 .........(2)

Verification of B:

3x 1 + 6x 2 = 30 ..........[(1) × 3] 3x 1 + 4x 2 = 24 .........(2) −     −       −        2x 2 = 6

From (1), x 1 + 6 = 10

∴ B is (4, 3) Minimum value occurs at B(4, 3)

∴ The solution is x 1 = 4, x 2 = 3 and Z min = 2300. • Maharashtra Board Question Bank with Solutions (Official)
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