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Question
Solve the following LPP:
Maximize z = 6x + 10y subject to 3x + 5y ≤ 10, 5x + 3y ≤ 15, x ≥ 0, y ≥ 0.
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Solution
First we draw the lines AB and CD whose equations are 3x + 5y = 10 and 5x + 3y = 15 respectively.
| Line | Equation | Points on the X-axis |
Points on the Y-axis |
Sign | Region |
| AB | 3x + 5y = 10 | A`(10/3, 0)` | B(0, 2) | ≤ | origin side of line AB |
| CD | 5x + 3y = 15 | C(3, 0) | D(0, 5) | ≤ | origin side of line CD |
The feasible region is OCPBD which is shaded region in the graph.
The vertices of the feasible region are O(0, 0), C(3, 0), P and B(0, 2).
P is the point of intersection of lines
3x + 5y = 10 ....(1)
and 5x + 3y = 15 ....(2)
Multiplying equation (1) by 5 and equation (2) by 3, we get
15x + 25y = 50
15x + 9y = 45
On subtracting, we get
16y = 5
∴ y = `5/16`
Substituting y = `5/16` in equation (1), we get
`3x + 25/16` = 10
∴ 3x = `10 - 25/16 = 135/16`
∴ x = `45/16`
∴ P ≡ `(45/16, 5/16)`
The values of objective function z = 6x + 10y at these vertices are
z(O) = 6(0) + 10(0)
= 0 + 0
= 0
z(C) = 6(3) + 10(0)
= 18 + 0
= 18
z(P) = `6(45/16) + 10(5/10)`
= `270/16 + 50/16`
= `320/16`
= 20
z(B) = 6(0) + 10(2)
= 0 + 20
= 20
The maximum value of z is 20 at P`(45/16, 5/16)` and B(0, 2) two consecutive vertices.
∴ z has maximum value 20 at each point of line segment
PB where B is (0, 2) and P is `(45/16, 5/16)`
Hence, there are infinite number of optimum solutions.
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