Question

The difference between the maximum value and the minimum value of the objective function \(z = 3x+y\) subject to the constraints \(2x+3y\leq 6\), \(x+y\geq 1\), \(x\geq 0\), \(y\geq 0\) is....

• A. \(7\)
• B. \(3\)
• C. \(8\)
• D. \(1\)

Hint:

Find the corner points of the feasible region and evaluate z at each.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
The optimum of a linear objective function over a bounded feasible region occurs at a corner point.

Step 2: Feasible region:
The constraints are \(2x + 3y \le 6\), \(x + y \ge 1\), \(x \ge 0\), \(y \ge 0\). Corner points:
\((1, 0)\) from \(x + y = 1\) and \(y = 0\); \((3, 0)\) from \(2x + 3y = 6\) and \(y = 0\); \((0, 2)\) from \(2x+3y=6\) and \(x = 0\); \((0, 1)\) from \(x + y = 1\) and \(x = 0\).

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

Step 4: Result:
Maximum is 9 and minimum is 1, so the difference is \(9 - 1 = 8\).

Step 5: Why the other options are wrong.
Option 7 would be \(9 - 2\), taking the wrong minimum. Option 3 is \(3 - 0\)-type slips, and 1 is the minimum itself.

Final Answer:
The difference is 8, option (C). \[ \boxed{8} \]