Question

Consider the following regions: \[ S_1 = \{(x_1, x_2) \in \mathbb{R}^2 : 2x_1 + x_2 \leq 4, \quad x_1 + 2x_2 \leq 5, \quad x_1, x_2 \geq 0\} \] \[ S_2 = \{(x_1, x_2) \in \mathbb{R}^2 : 2x_1 - x_2 \leq 5, \quad x_1 + 2x_2 \leq 5, \quad x_1, x_2 \geq 0\} \] Then, which of the following is/are TRUE?

• A. The maximum value of \( x_1 + x_2 \) is 3 on the region \( S_2 \)
• B. The maximum value of \( x_1 + x_2 \) is 5 on the region \( S_2 - S_1 \)
• C. The maximum value of \( x_1 + x_2 \) is 3 on the region \( S_1 \cap S_2 \)
• D. The maximum value of \( x_1 + x_2 \) is 4 on the region \( S_1 \cup S_2 \)

Hint:

To find the maximum value of \( x_1 + x_2 \) in a region defined by inequalities, check the boundary lines and evaluate at the intersection points.

Answer 1

The Correct Option is C, D

Step 1: Graphing the Regions \( S_1 \) and \( S_2 \)
First, we need to graph the constraints defining the regions \( S_1 \) and \( S_2 \). The inequalities can be rewritten as linear equations to find the boundary lines:
- For \( S_1 \), the boundaries are given by:
\[ 2x_1 + x_2 = 4 \quad \text{and} \quad x_1 + 2x_2 = 5 \]
- For \( S_2 \), the boundaries are given by:
\[ 2x_1 - x_2 = 5 \quad \text{and} \quad x_1 + 2x_2 = 5 \]
These boundaries define the respective feasible regions in the first quadrant \( x_1, x_2 \geq 0 \).

Step 2: Finding the Intersection of \( S_1 \) and \( S_2 \)
We now find the intersection of the regions \( S_1 \) and \( S_2 \), which is represented by \( S_1 \cap S_2 \). To do this, we solve the system of equations:
\[ 2x_1 + x_2 = 4 \quad \text{and} \quad x_1 + 2x_2 = 5 \]
From the first equation, solve for \( x_2 \):
\[ x_2 = 4 - 2x_1 \]
Substitute this into the second equation:
\[ x_1 + 2(4 - 2x_1) = 5 \quad \Rightarrow \quad x_1 + 8 - 4x_1 = 5 \quad \Rightarrow \quad -3x_1 = -3 \quad \Rightarrow \quad x_1 = 1 \]
Substitute \( x_1 = 1 \) into \( x_2 = 4 - 2x_1 \):
\[ x_2 = 4 - 2(1) = 2 \]
Thus, the intersection point is \( (1, 2) \).

Step 3: Evaluating the Maximum Value of \( x_1 + x_2 \) in \( S_1 \cap S_2 \)
At the intersection point \( (1, 2) \), we calculate the value of \( x_1 + x_2 \):
\[ x_1 + x_2 = 1 + 2 = 3 \]
Thus, the maximum value of \( x_1 + x_2 \) on the region \( S_1 \cap S_2 \) is \( 3 \).

Step 4: Finding the Maximum Value of \( x_1 + x_2 \) on \( S_1 \cup S_2 \)
The maximum value of \( x_1 + x_2 \) on the union of \( S_1 \) and \( S_2 \) occurs at the boundary points. We evaluate \( x_1 + x_2 \) on the boundary of both regions.
- From the boundary of \( S_1 \) at \( 2x_1 + x_2 = 4 \), the maximum value occurs at \( x_1 = 2, x_2 = 0 \), where \( x_1 + x_2 = 2 + 0 = 2 \).
- From the boundary of \( S_2 \), particularly \( x_1 = 4, x_2 = 0.5 \) (from solving \( 2x_1 - x_2 = 5 \) and \( x_1 + 2x_2 = 5 \)), you can achieve a maximum \( x_1 + x_2 = 4 \).
Thus, the maximum value of \( x_1 + x_2 \) on the region \( S_1 \cup S_2 \) is \( 4 \).

\[ \boxed{C} \quad \text{The maximum value of } x_1 + x_2 \text{ is 3 on the region } S_1 \cap S_2. \]
\[ \boxed{D} \quad \text{The maximum value of } x_1 + x_2 \text{ is 4 on the region } S_1 \cup S_2. \]