Question

The solution set for minimizing the function \( z = x + y \) with constraints \[ x + y \geq 2, \quad x + 2y \leq 8, \quad x \leq 3, \quad x, y \geq 0 \] contains

• A. \( x = 0, y = 3 \)
• B. \( x = 8, y = 0 \)
• C. infinitely many points
• D. \( x = 2, y = 3 \)

Hint:

To minimize a linear function subject to linear constraints, graphically plot the constraints and identify the feasible region. The minimum value is found at the vertex of this region.

Answer 1

The Correct Option is A

Step 1: Understanding the given function and constraints.
We are asked to minimize the function \( z = x + y \) subject to the constraints:
\[ x + y \geq 2, \quad x + 2y \leq 8, \quad x \leq 3, \quad x, y \geq 0. \]
We will use the method of graphical analysis to solve this problem by plotting the constraints and finding the feasible region.

Step 2: Plotting the constraints.
We start by plotting the constraint \( x + y \geq 2 \), which represents the region to the right of the line \( x + y = 2 \). This line has a slope of \( -1 \) and passes through the point \( (2, 0) \) and \( (0, 2) \).
Next, we plot the constraint \( x + 2y \leq 8 \), which represents the region below the line \( x + 2y = 8 \). This line has a slope of \( -\frac{1}{2} \) and passes through the point \( (8, 0) \) and \( (0, 4) \).
The third constraint \( x \leq 3 \) is a vertical line at \( x = 3 \), and the fourth constraint \( x, y \geq 0 \) restricts the solution to the first quadrant.

Step 3: Identifying the feasible region.
The feasible region is the intersection of all the above constraints. We need to find the points of intersection of the lines and check if they satisfy all the constraints.

Step 4: Finding the intersection points.
We solve the system of equations formed by the lines \( x + y = 2 \) and \( x + 2y = 8 \). Solving this system, we subtract the first equation from the second:
\[ (x + 2y) - (x + y) = 8 - 2, \] \[ y = 6. \]
Substitute \( y = 6 \) into \( x + y = 2 \):
\[ x + 6 = 2 \quad \Rightarrow \quad x = -4. \]
Since \( x = -4 \) does not satisfy the constraint \( x \geq 0 \), this point is not feasible.
Now, consider the intersection of \( x + y = 2 \) and \( x = 3 \). Substituting \( x = 3 \) into \( x + y = 2 \):
\[ 3 + y = 2 \quad \Rightarrow \quad y = -1. \]
Since \( y = -1 \) does not satisfy \( y \geq 0 \), this point is also not feasible.

Step 5: Conclusion.
The only feasible point that satisfies all the constraints is \( x = 0, y = 3 \).
Final Answer:
Thus, the correct answer is:
\[ \boxed{x = 0, y = 3}. \]