Consider the following Linear programming problem (LPP): Maximize…
2019
Consider the following Linear programming problem (LPP):
Maximize \(𝑧=𝑥_1+𝑥_2\)
Subject to the constraints:
\(𝑥_1+2𝑥_2≤2000 \\ 𝑥_1+𝑥_2≤1500 \\ 𝑥_2≤600 \\ and \ 𝑥_1,𝑥_2≥0\)
The solution of the above LPP is
- A.
\(x_1=750, x_2= 750, z=1500\) - B.
\(x_1=500, x_2= 1000, z=1500\) - C.
\(x_1=1000, x_2= 500, z=1500\) - D.
\(x_1=900, x_2= 600, z=1500\)
Attempted by 7 students.
Show answer & explanation
Correct answer: C
Key insight: From the constraint x1 + x2 ≤ 1500 we get z = x1 + x2 ≤ 1500, so the objective cannot exceed 1500.
Find the intersection of x1 + x2 = 1500 and x1 + 2x2 = 2000. Subtracting the first from the second gives x2 = 500, so x1 = 1000.
Check feasibility of (1000, 500): x1 + 2x2 = 1000 + 2·500 = 2000 ≤ 2000, x1 + x2 = 1500 ≤ 1500, and x2 = 500 ≤ 600. So (1000, 500) is feasible.
Check another extreme: x2 = 0 with x1 + x2 = 1500 gives (1500, 0), which also satisfies x1 + 2x2 ≤ 2000 and gives z = 1500.
For any t in [0, 500], the point (x1, x2) = (1500 − t, t) satisfies all constraints (because x2 ≤ 500 ensures x1 + 2x2 ≤ 2000) and gives z = 1500.
Conclusion: The maximum value is z = 1500. The point (1000, 500) is a valid optimal solution (it lies at the intersection of x1 + x2 = 1500 and x1 + 2x2 = 2000). There are infinitely many optimal solutions along the segment between (1500, 0) and (1000, 500), all giving z = 1500.