Q.Maximize Subject to constraints:
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Start your 14-day free trial to unlock the full solution →This is a linear programming problem where the constraints are contradictory — the region and cannot both hold for non-negative , so the feasible region is empty and the problem has no solution.
The first thing to notice is that we're being asked to maximize a linear function over a region defined by linear inequalities. That's a classic Linear Programming (LP) problem. In LP, if the feasible region (the set of all points satisfying all constraints) is non-empty and bounded, the maximum occurs at a corner point. If the region is unbounded, the maximum might be infinite — or might not exist. But there's a third possibility: the constraints might be inconsistent, meaning no point satisfies all of them at once. That's what happens here.
Let's see why.
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Rewrite the constraints in a clearer form.
We have:
- , (non-negativity)
The second constraint can be rewritten as .
The third constraint says is at least as large as .
So we need a point where simultaneously:
- Combine the two inequalities. If and , then by transitivity:
This forces , which simplifies to , an impossibility.
No real numbers can satisfy both conditions at the same time. The two inequalities are contradictory.
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Check if non-negativity changes anything.
Even if we ignore , the contradiction is absolute — it doesn't depend on sign. So the non-negativity constraints don't help; they only make the region smaller, not larger.
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Conclusion about the feasible region.
Since no point can satisfy all constraints, the feasible region is empty. In linear programming, when the feasible region is empty, the problem is said to have no feasible solution. There is nothing to maximize. …
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