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: Use the objective's direction:
Since $z = 4x + y$ weights $x$ much more than $y$, the optimum is near the largest feasible $x$.

Step 2: Largest x:
On the x-axis the constraints give $x \le 5$, $x \le 3.5$, $x \le \frac{11}{3}$. The strictest is $x \le 3.5$, so $(3.5, 0)$ is feasible and gives $z = 14$.

Step 3: Check that nothing beats it:
Moving up the line $2x + y = 7$ from $(3.5, 0)$, $z = 4x + (7-2x) = 2x + 7$ decreases as $x$ falls, and at $(3,1)$ it is 13. The remaining corners give 5 and 8.

Final Answer:
Maximum $z = 14$, option (D). \[ \boxed{14 \text{ (D)}} \]