Question:

The maximum value of \(Z=6x+3y\) subject to \(x+y\le 25,\ x\ge0,\ y\ge0\) is

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Two-variable LPP: just test $Z$ at the corner points.
  • \(150\)
  • \(225\)
  • \(425\)
  • none of these
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The Correct Option is A

Solution and Explanation

The feasible region is the triangle with vertices where the lines meet: \((0,0)\), \((25,0)\), \((0,25)\). A linear objective hits its max at a vertex. Evaluate \(Z\): \((25,0)\Rightarrow Z=6\cdot25+3\cdot0=150\). \((0,25)\Rightarrow Z=0+75=75\). \((0,0)\Rightarrow Z=0\). So the maximum is \(150\) at \((25,0)\).
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