For the following LPP Maximise Z = 3x + 4y subject to constraints x – y ≥ – 1, x ≤ 3 x ≥ 0, y ≥ 0 The maximum value is
Constraints in linear programming are conditions or limitations that restrict the values of variables in the optimization problem. They are typically expressed as linear inequalities or equations. Constraints define the feasible region within which the objective function is optimized, ensuring that the solution meets specified requirements or limitations.
Linear Programming is a method used to optimize a linear objective function subject to linear constraints. It involves finding the feasible region formed by these constraints. The optimal solution lies at the vertices of the feasible region. Linear Programming is applied in various fields such as economics business and resource management to make optimal decisions.
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To solve the LPP, we first plot the constraints:
1. x – y ≥ -1 ⟹ y ≤ x + 1
2. x ≤ 3
3. x ≥ 0
4. y ≥ 0
Now, we find the corner points of the feasible region:
– From x = 0 and y = 0, the point is (0, 0).
– For the line x = 3 and y = x + 1, when x = 3, y = 4. So the point is (3, 4).
– From the intersection of x = 3 and y = x + 1, we get (3, 4).
Now, evaluate Z = 3x + 4y at the corner points:
– At (0, 0), Z = 3(0) + 4(0) = 0
– At (3, 4), Z = 3(3) + 4(4) = 9 + 16 = 25
The maximum value of Z occurs at (3, 4) and is 25.
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