Question:
The maximum value of \(Z = 4x+5y\), subject to the constraints \(3x+y\leq 15,3x+4y\leq 24,x\geq 0,y\geq 0\) is
The maximum value of \(Z = 4x+5y\), subject to the constraints \(3x+y\leq 15,3x+4y\leq 24,x\geq 0,y\geq 0\) is
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Check Z at each corner point of the feasible region.
Updated On: Oct 1, 2026
- \(31\)
- \(30\)
- \(42\)
- \(47\)
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The Correct Option is A
Solution and Explanation
Step 1: Understanding the Concept:
The maximum of a linear function over a bounded feasible region occurs at a corner point. Find all corner points of the region defined by the constraints.
Step 2: Key Formula or Approach:
Corner points come from the axes and from the intersection of \(3x + y = 15\) and \(3x + 4y = 24\).
Step 3: Detailed Explanation:
Subtract the equations: \(3y = 9\), so \(y = 3\) and \(x = 4\). The point is \((4, 3)\).
Other corners: \((0,0)\), \((5, 0)\) (from \(3x = 15\)) and \((0, 6)\) (from \(4y = 24\)).
\(Z(0,0) = 0\), \(Z(5, 0) = 20\), \(Z(0, 6) = 30\), \(Z(4, 3) = 16 + 15 = 31\).
\[ Z_{max} = 31 \]
Final Answer:
The maximum value is \(31\), option (A). \[ \boxed{31} \]
The maximum of a linear function over a bounded feasible region occurs at a corner point. Find all corner points of the region defined by the constraints.
Step 2: Key Formula or Approach:
Corner points come from the axes and from the intersection of \(3x + y = 15\) and \(3x + 4y = 24\).
Step 3: Detailed Explanation:
Subtract the equations: \(3y = 9\), so \(y = 3\) and \(x = 4\). The point is \((4, 3)\).
Other corners: \((0,0)\), \((5, 0)\) (from \(3x = 15\)) and \((0, 6)\) (from \(4y = 24\)).
\(Z(0,0) = 0\), \(Z(5, 0) = 20\), \(Z(0, 6) = 30\), \(Z(4, 3) = 16 + 15 = 31\).
\[ Z_{max} = 31 \]
Final Answer:
The maximum value is \(31\), option (A). \[ \boxed{31} \]
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