Question

Q. 28. Solve the following linear programming problem:
Maximise : = 150x + 250 y
Subject to: 4x + y = 40
3x + 2y =60
x=0
y=0​

Answer 1

Maximum value is 7500

Given:

      Maximise : = 150x + 250 y

      Subject to: 4x + y = 40

                        3x + 2y =60

                               x=0

                               y=0​

Find:

Maximum value of graph.

Explanation:

• First equation constrain are given by.
• 4x + y = 40
• Put y = 0 and find value of x.
• 4x + 0 = 40
• x=$\frac{40}{4}$ =10  first point (10,0)
• Put x = 0 and find value of y.
• y = 40     second point (0,40)
• Similarly for second equation put value of x and y = 0.
• Therefore 3x + 2y =60
• 3x = 60 → x =20 point = (20,0)
• Similarly, 2y =60 → y= 30  point (0,30)
• Constrain of the equation  4x + y = 40 is (10,0) and (0,40)
• Similarly, for equation 3x + 2y =60 is (20,0) and (0,30)
• x is grater than 0 and y is grater than 0
• Draw the graph of above constrains.
• From graph the bonded region for this contain is (0,30) (4,24) and (10,0)
• From this we can check  maximum value for each point.
• For (0,30) Z= 150 x + 250 y  = 150×0 + 250×30 = 7,500
• For (4,24) Z = 150×4 + 250×24 = 6,600
• For (10,0) Z = 150×10 + 250×0 = 1500
• Out of the above equation maximum value is 7,500.

Hence maximum value is 7500.

To learn more...

1 How to convert maximization to minimization in assignment problem?

https://brainly.in/question/7377811

2 Give a general formulation of linear programming problem of cost maximization ​

https://brainly.in/question/9606509