solve the following linear programming problem graphically minimize z=200x500y

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Example 2 - Chapter 12 Class 12 Linear Programming

Last updated at May 29, 2023 by Teachoo

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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

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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.

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