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Linear Programming

Question
CBSEENMA12033528
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Solve the following Linear Programming Problems graphically:
Minimise    Z = 5x + 3y
subject to the constraints: 3x + 5y ≤ 15, 5x + 2y ≤ 10, x ≥ 0, y ≥ 0.

Solution

We have to maximise
Z = 5x + 3 y
subject to the constraints
3x + 5 y ≤ 15
5x + 2 y ≤ 10
x ≥ 0, y ≥ 0
Consider a set of rectangular cartesian axes OXY in the plane.
It is clear that any point which satisfies x ≥ 0, y ≥ 0 lies in the first quadrant.
Let us draw the graph of 3x + 5 y= 15
For x = 0, 5 y = 15 or y = 3
For y = 0, 3x = 15 or x = 5
∴ line meets OX in A(5, 0) and OY in L(0, 3).
Again we draw the graph of 5x + 2 y = 10
For x = 0, 2 y = 10 or y = 5
For y = 0, 5x = 10 or x = 2
∴ line meets OX in B(2, 0) and OY in M(0, 5).
Since feasible region is the region which satisfies all the constraints.

∴ OBCL is the feasible region and corner points are O(0, 0), B(2, 0),
    straight C open parentheses 20 over 19 comma space 45 over 19 close parentheses comma space space space space straight L left parenthesis 0 comma space 3 right parenthesis.
At straight O left parenthesis 0 comma space 0 right parenthesis comma space                                straight Z space equals space 0 plus 0 space equals space 0
At straight B left parenthesis 2 comma space 0 right parenthesis comma                                 straight Z space equals 10 plus 0 space equals 10
At  straight C open parentheses 20 over 19 comma space 45 over 19 close parentheses comma space space space straight Z space equals 100 over 19 plus 135 over 19 space equals space 235 over 19
At space space space space straight L left parenthesis 0 comma space 3 right parenthesis comma space space space space space space space straight Z space equals space 0 plus 9 space equals space 9
therefore space space space maximum space value space space equals space 235 over 19 space at space open parentheses 20 over 19 comma space 45 over 19 close parentheses.

Some More Questions From Linear Programming Chapter

Minimize z = 2x + 3y, such that 1 ≤ x + 2y ≤ 10, x ≥ 0, y ≥ 0.

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Minimise and Maximise Z = 3x + 9y
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Minimise and Maximise Z = x + 2y subject to constraints x + 2y ≥ 100, 2x - y ≤ 0, 2x + y ≤ 200, x, y ≥ 0

 

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