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:
This is a Linear Programming Problem (LPP). We are tasked with maximizing the linear objective function $z = 10x + 25y$ given a system of boundary linear inequalities.

Step 2: Key Formula or Approach:
According to the Corner Point Theorem, the maximum or minimum value of an objective function in a bounded feasible region always occurs at one of the corner vertices of that region. 1. Identify the corner vertices of the feasible region defined by the constraints. 2. Substitute each vertex into $z$ to locate the maximum value.

Step 3: Detailed Explanation:
Let's find the boundaries of the closed feasible region: $0 \le x \le 3$ forms vertical lines at $x=0$ and $x=3$. $0 \le y \le 3$ forms horizontal lines at $y=0$ and $y=3$. $x + y \le 5$ is a diagonal line cutting across the upper bound boundaries. The intersection points forming the vertices of this bounded polygon are: 1. Origin: $O(0, 0)$ 2. Along the x-axis up to its constraint line: $A(3, 0)$ 3. Intersection of $x = 3$ and $x + y = 5 \implies y = 2$, yielding $B(3, 2)$. 4. Intersection of $y = 3$ and $x + y = 5 \implies x = 2$, yielding $C(2, 3)$. 5. Along the y-axis up to its constraint line: $D(0, 3)$ Let's evaluate the objective function $z = 10x + 25y$ at these corner points: At $O(0, 0)$: $z = 10(0) + 25(0) = 0$ At $A(3, 0)$: $z = 10(3) + 25(0) = 30$ At $B(3, 2)$: $z = 10(3) + 25(2) = 30 + 50 = 80$ At $C(2, 3)$: $z = 10(2) + 25(3) = 20 + 75 = 95$ At $D(0, 3)$: $z = 10(0) + 25(3) = 75$ Comparing the values, the absolute maximum value is $95$, which occurs exactly at the coordinate point $(2, 3)$.

Step 4: Final Answer:
The maximum value occurs at the point $(2, 3)$, which corresponds to option (B).