Question

The minimum value of \(Z = 3x+y\), subject to the constraints \(2x+3y\leq 6,x+y\geq 1,x\geq 0,y\geq 0\) is....

• A. \(5\)
• B. \(2\)
• C. \(1\)
• D. \(9\)

Hint:

Evaluate Z at all corners of the feasible region.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
In graphical LPP, the optimal value of \(Z\) over a bounded feasible region lies at a corner point.

Step 2: Draw the constraints:
\(2x+3y\le6\) is the region below the line through \((3,0)\) and \((0,2)\). \(x+y\ge1\) is the region above the line through \((1,0)\) and \((0,1)\). With \(x,y\ge0\) we are in the first quadrant.

Step 3: Corner points:
The feasible region is a quadrilateral with corners \((1,0)\), \((3,0)\), \((0,2)\) and \((0,1)\).

Step 4: Evaluate Z = 3x + y:
\((1,0)\): \(3\). \((3,0)\): \(9\). \((0,2)\): \(2\). \((0,1)\): \(1\).

Step 5: Choose:
The minimum value is \(1\) at \((0,1)\), option (C). The value \(9\) is the maximum.

Final Answer:
The minimum of Z is 1 at (0, 1). \[ \boxed{1} \]