Question

The maximum value of $z = 10x + 25y$ subject to $0 \le x \le 3$, $0 \le y \le 3$, $x + y \le 5$ occurs at the point

• A. $(3, 2)$
• B. $(2, 3)$
• C. $(4, 3)$
• D. $(5, 4)$

Hint:

Notice the weighting in the objective function $z = 10x + 25y$. The coefficient for $y$ ($25$) is much larger than the coefficient for $x$ ($10$). To maximize $z$, you want to make $y$ as large as possible first. The maximum allowed value for $y$ is $3$. Plugging $y = 3$ into the diagonal constraint $x + y \le 5$ gives $x = 2$, leading you straight to the optimal point $(2, 3)$ without checking all vertices!

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
Maximize z = 10x + 25y subject to 0 ≤ x ≤ 3, 0 ≤ y ≤ 3, x + y ≤ 5.

Step 2: Key Formula or Approach:
By the Corner Point Theorem, evaluate z at each vertex of the feasible region. The maximum occurs at one of these corner points.

Step 3: Detailed Explanation:
Vertices: O(0,0) → z=0; A(3,0) → z=30; B(3,2) [from x=3, x+y=5] → z=10(3)+25(2)=80; C(2,3) [from y=3, x+y=5] → z=10(2)+25(3)=95; D(0,3) → z=75. Maximum is 95 at (2,3).

Step 4: Final Answer:
Maximum occurs at (2, 3), option (B).