Question

The maximum value of \(z = 4x+y\) subject to the constraints \(x+y\leq 5,2x+y\leq 7,3x+2y\leq 11,x\geq 0,y\geq 0\) is \(\ldots\)

• A. \(13\)
• B. \(8\)
• C. \(11\)
• D. \(14\)

Hint:

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

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
In a linear programming problem the maximum of a linear function occurs at a corner point of the feasible region.

Step 2: Key Formula or Approach:
Constraints: \(x + y \le 5\), \(2x + y \le 7\), \(3x + 2y \le 11\), \(x, y \ge 0\). Find the corners and test them in all constraints.

Step 3: Detailed Explanation:
Corner points:
\((0, 0)\): feasible, \(z = 0\).
\((3.5, 0)\) from \(2x + y = 7\) on the x-axis: check \(3(3.5) = 10.5 \le 11\) and \(3.5 \le 5\). Feasible, \(z = 4(3.5) = 14\).
\((0, 5)\) from \(x + y = 5\) on the y-axis: check \(2(5) = 10 \le 11\) and \(5 \le 7\). Feasible, \(z = 5\).
\((3, 1)\) from \(2x + y = 7\) and \(3x + 2y = 11\): check \(x + y = 4 \le 5\). Feasible, \(z = 13\).
\((1, 4)\) from \(x + y = 5\) and \(3x + 2y = 11\): check \(2x + y = 6 \le 7\). Feasible, \(z = 8\).
The corner \((2, 3)\) from \(x+y = 5\) and \(2x + y = 7\) fails \(3x + 2y = 12 > 11\), so it is not a vertex.
The largest value is 14 at \((3.5, 0)\).

Final Answer:
The maximum value of \(z\) is 14, option (D). \[ \boxed{14 \text{ (D)}} \]