Step 1: Understanding the Concept:
Linear Programming Problems (LPP) focus on optimizing a linear objective function within a set of linear constraints.
A fundamental principle of LPP is that the optimal solution (both maximum and minimum) always occurs at one of the vertices (corner points) of the feasible region.
The feasible region is the area on a coordinate plane where all the inequality constraints are satisfied simultaneously.
In this problem, we need to define the boundary lines, find their intersections to identify the corner points, evaluate the objective function \( Z \) at each point, and then use the difference to solve for \( a \).
Step 2: Key Formula or Approach:
The boundary lines are obtained by converting the inequalities into equations:
Line 1: \( x + y = 1 \)
Line 2: \( 5x + 10y = 50 \) (which can be simplified by dividing by 5: \( x + 2y = 10 \))
The constraints \( x, y \geq 0 \) restrict the feasible region to the first quadrant.
We will find the x and y intercepts for each line to plot them and find the vertices.
Step 3: Detailed Explanation:
Let's find the intercepts for the boundary lines.
For Line 1: \( x + y = 1 \)
If \( x = 0 \), then \( y = 1 \). Point is \( (0, 1) \).
If \( y = 0 \), then \( x = 1 \). Point is \( (1, 0) \).
For Line 2: \( x + 2y = 10 \)
If \( x = 0 \), then \( 2y = 10 \implies y = 5 \). Point is \( (0, 5) \).
If \( y = 0 \), then \( x = 10 \). Point is \( (10, 0) \).
The region is defined by \( x + y \geq 1 \) (outside the line relative to origin) and \( x + 2y \leq 10 \) (inside the line relative to origin).
Combined with \( x, y \geq 0 \), the vertices of the feasible region are:
\( A(1, 0), B(10, 0), C(0, 5), \) and \( D(0, 1) \).
Now, we evaluate the objective function \( Z = 3x + 5y \) at each of these four corner points:
1. At \( A(1, 0) \): \( Z = 3(1) + 5(0) = 3 \)
2. At \( B(10, 0) \): \( Z = 3(10) + 5(0) = 30 \)
3. At \( C(0, 5) \): \( Z = 3(0) + 5(5) = 25 \)
4. At \( D(0, 1) \): \( Z = 3(0) + 5(1) = 5 \)
From these evaluations, we identify the extremes:
Maximum value \( Z_{\text{max}} = 30 \) at point \( B \).
Minimum value \( Z_{\text{min}} = 3 \) at point \( A \).
The problem states that the difference between the maximum and minimum values is \( 3a \).
Calculation of the difference:
\[ Z_{\text{max}} - Z_{\text{min}} = 30 - 3 = 27 \]
Setting this equal to the given expression:
\[ 3a = 27 \]
Dividing both sides by 3:
\[ a = 9 \]
Thus, the value of \( a \) is 9.
Step 4: Final Answer:
By identifying the vertices of the feasible region and calculating the objective function's values at those points, we found the range to be 27. Equating this to \( 3a \) gave \( a = 9 \). This matches Option (C).