Question

Consider the Linear Programming Problem: Maximize $Z=x+2y$ Subject to $2x+3y\le12, x\ge0, y\ge0$. The optimal value is} \textit{Note: A stray '0.' from the original document has been removed for clarity.

• A. 3
• B. 7
• C. 8
• D. 10
• E. 12

Hint:

Logic Tip: You can also compare the slope of the objective function line ($m_Z = -1/2$) to the slope of the constraint line ($m_c = -2/3$). Because the objective function is steeper relative to the y-axis ($1/2<2/3$ when viewing positive magnitudes), the maximum will lie on the y-intercept.

Answer 1

The Correct Option is C

Concept:
The fundamental theorem of linear programming states that the optimal value of the objective function (max or min) always occurs at one of the corner points (vertices) of the feasible region bounded by the constraints.
Step 1: Identify the feasible region constraints.
The primary constraint is the line: $2x + 3y = 12$ The non-negativity constraints $x \ge 0$ and $y \ge 0$ restrict the region to the first quadrant. Since the inequality is $\le 12$, the feasible region is the triangle bounded by the x-axis, y-axis, and the line.
Step 2: Find the corner points (vertices) of the region.
We need the intersection points of the boundary lines. 1. Intersection of $x=0$ and $y=0$ (The origin): Point A: $(0, 0)$ 2. Intersection of the line and the x-axis (Set $y=0$): $2x + 3(0) = 12 \implies 2x = 12 \implies x = 6$ Point B: $(6, 0)$ 3. Intersection of the line and the y-axis (Set $x=0$): $2(0) + 3y = 12 \implies 3y = 12 \implies y = 4$ Point C: $(0, 4)$
Step 3: Evaluate the objective function Z at each corner point.
The objective function is $Z = x + 2y$. At Point A $(0, 0)$: $$Z = 0 + 2(0) = 0$$ At Point B $(6, 0)$: $$Z = 6 + 2(0) = 6$$ At Point C $(0, 4)$: $$Z = 0 + 2(4) = 8$$
Step 4: Identify the maximum optimal value.
Comparing the results ($0, 6, 8$), the maximum value is $8$.