Question

The maximum value of the objective function $z = 2x + 3y$ subject to the constraints $x + y \le 5$, $2x + y \ge 4$ and $x \ge 0, y \ge 0$ is

• A. 15
• B. 10
• C. 20
• D. 25

Hint:

Look closely at the objective function $z = 2x + 3y$. The weighting coefficient for $y$ ($3$) is larger than for $x$ ($2$). To maximize $z$ quickly when the constraints allow, test the points that lie highest along the vertical $y$-axis first. Comparing $(0,5)$ and $(0,4)$ immediately yields the maximum value of $15$.

Answer 1

The Correct Option is A

Step 1: State the method.
For an LPP, the maximum of $z = 2x + 3y$ over a feasible region occurs at a corner point, so we find the corners.
Step 2: Draw the boundary lines.
$x + y = 5$ joins $(5,0)$ and $(0,5)$; $2x + y = 4$ joins $(2,0)$ and $(0,4)$. The region also stays in the first quadrant.
Step 3: Identify the corner points.
The feasible vertices are $A(0,4)$, $B(2,0)$, $C(5,0)$ and $D(0,5)$.
Step 4: Evaluate z at A and B.
$z(A) = 2(0)+3(4) = 12$; $z(B) = 2(2)+3(0) = 4$.
Step 5: Evaluate z at C and D.
$z(C) = 2(5)+3(0) = 10$; $z(D) = 2(0)+3(5) = 15$.
Step 6: Pick the largest.
The biggest value is $15$ at $D(0,5)$, which is option 1 and matches the key.
\[ \boxed{z_{\max} = 15} \]