Answer :
To solve the given Linear Programming Problem (L.P.P.) and determine the range of values for [tex]\( C_2 \)[/tex] that keeps the current solution optimal, we need to follow the steps below:
### Step 1: Identify the constraints and the objective function
The given L.P.P. is:
[tex]\[ Z_{\max } = x_1 + \frac{3}{2} x_2 \][/tex]
subject to:
[tex]\[ x_1 + x_2 \leq 80 \quad \text{(1)} \][/tex]
[tex]\[ x_1 + 2 x_2 \leq 120 \quad \text{(2)} \][/tex]
[tex]\[ 2 x_1 + x_2 \leq 140 \quad \text{(3)} \][/tex]
with
[tex]\[ x_1, x_2 \geq 0. \][/tex]
### Step 2: Determine the feasible region
We need to determine the intersection points of the constraint lines. Solving pairs of linear equations will provide the vertices of the feasible region.
#### Intersection of Constraints (1) and (2):
[tex]\[ \begin{cases} x_1 + x_2 = 80 \\ x_1 + 2 x_2 = 120 \end{cases} \][/tex]
Subtracting the first equation from the second:
[tex]\[ (x_1 + 2 x_2) - (x_1 + x_2) = 120 - 80 \\ x_2 = 40 \\ x_1 + 40 = 80 \\ x_1 = 40 \][/tex]
So, the intersection point is [tex]\( (40, 40) \)[/tex].
#### Intersection of Constraints (1) and (3):
[tex]\[ \begin{cases} x_1 + x_2 = 80 \\ 2 x_1 + x_2 = 140 \end{cases} \][/tex]
Subtracting the first equation from the second:
[tex]\[ (2 x_1 + x_2) - (x_1 + x_2) = 140 - 80 \\ x_1 = 60 \\ 60 + x_2 = 80 \\ x_2 = 20 \][/tex]
So, the intersection point is [tex]\( (60, 20) \)[/tex].
#### Intersection of Constraints (2) and (3):
[tex]\[ \begin{cases} x_1 + 2 x_2 = 120 \\ 2 x_1 + x_2 = 140 \end{cases} \][/tex]
Multiplying the first equation by 2:
[tex]\[ 2 x_1 + 4 x_2 = 240 \\ 2 x_1 + x_2 = 140 \][/tex]
Subtracting the second from the first:
[tex]\[ (2 x_1 + 4 x_2) - (2 x_1 + x_2) = 240 - 140 \\ 3 x_2 = 100 \\ x_2 = \frac{100}{3} \approx 33.33 \][/tex]
Substitute [tex]\( x_2 \)[/tex] into the second equation:
[tex]\[ 2 x_1 + \frac{100}{3} = 140 \\ 2 x_1 = 140 - \frac{100}{3} \\ 2 x_1 = \frac{320}{3} \\ x_1 = \frac{160}{3} \approx 53.33 \][/tex]
So, the intersection point is [tex]\( \left(\frac{160}{3}, \frac{100}{3}\right) \)[/tex].
### Step 3: Evaluate the Objective Function at Vertices
Calculate [tex]\( Z_{\max} \)[/tex] at the vertices:
1. At [tex]\( (0, 0) \)[/tex]:
[tex]\[ Z = 0 + \frac{3}{2} \times 0 = 0 \][/tex]
2. At [tex]\( (40, 40) \)[/tex]:
[tex]\[ Z = 40 + \frac{3}{2} \times 40 = 40 + 60 = 100 \][/tex]
3. At [tex]\( (60, 20) \)[/tex]:
[tex]\[ Z = 60 + \frac{3}{2} \times 20 = 60 + 30 = 90 \][/tex]
4. At [tex]\( \left(\frac{160}{3}, \frac{100}{3}\right) \)[/tex]:
[tex]\[ Z = \frac{160}{3} + \frac{3}{2} \times \frac{100}{3} = \frac{160}{3} + \frac{150}{3} = \frac{310}{3} \approx 103.33 \][/tex]
### Step 4: Optimal Solution and Determination of [tex]\( C_2 \)[/tex]
The current optimal solution is at [tex]\( \left(\frac{160}{3}, \frac{100}{3}\right) \)[/tex] with [tex]\( Z = 103.33 \)[/tex].
When the objective function changes to:
[tex]\[ Z_{\max } = x1 + C_2 x2 \][/tex]
We need this vertex to remain optimal:
[tex]\[ Z = \left(\frac{160}{3}\right) + C_2 \left(\frac{100}{3}\right) = 103.33 \][/tex]
Therefore:
[tex]\[ \frac{160}{3} + C_2 \left(\frac{100}{3}\right) \leq 103.33 \][/tex]
To determine the exact range:
1. Rewrite the relation:
[tex]\[ \frac{160 + 100 C_2 }{3} = 103.33 \][/tex]
[tex]\[ 160 + 100 C_2 = 310 \][/tex]
[tex]\[ 100 C_2 = 150 \][/tex]
[tex]\[ C_2 = 1.5 \][/tex]
Thus, this indicates that for [tex]\( \left(\frac{160}{3}, \frac{100}{3}\right) \)[/tex] to remain optimal, [tex]\( C_2 \leq 1.5 \)[/tex].
So, the present solution will be optimal within the range of [tex]\( C_2 \leq 1.5 \)[/tex].
### Step 1: Identify the constraints and the objective function
The given L.P.P. is:
[tex]\[ Z_{\max } = x_1 + \frac{3}{2} x_2 \][/tex]
subject to:
[tex]\[ x_1 + x_2 \leq 80 \quad \text{(1)} \][/tex]
[tex]\[ x_1 + 2 x_2 \leq 120 \quad \text{(2)} \][/tex]
[tex]\[ 2 x_1 + x_2 \leq 140 \quad \text{(3)} \][/tex]
with
[tex]\[ x_1, x_2 \geq 0. \][/tex]
### Step 2: Determine the feasible region
We need to determine the intersection points of the constraint lines. Solving pairs of linear equations will provide the vertices of the feasible region.
#### Intersection of Constraints (1) and (2):
[tex]\[ \begin{cases} x_1 + x_2 = 80 \\ x_1 + 2 x_2 = 120 \end{cases} \][/tex]
Subtracting the first equation from the second:
[tex]\[ (x_1 + 2 x_2) - (x_1 + x_2) = 120 - 80 \\ x_2 = 40 \\ x_1 + 40 = 80 \\ x_1 = 40 \][/tex]
So, the intersection point is [tex]\( (40, 40) \)[/tex].
#### Intersection of Constraints (1) and (3):
[tex]\[ \begin{cases} x_1 + x_2 = 80 \\ 2 x_1 + x_2 = 140 \end{cases} \][/tex]
Subtracting the first equation from the second:
[tex]\[ (2 x_1 + x_2) - (x_1 + x_2) = 140 - 80 \\ x_1 = 60 \\ 60 + x_2 = 80 \\ x_2 = 20 \][/tex]
So, the intersection point is [tex]\( (60, 20) \)[/tex].
#### Intersection of Constraints (2) and (3):
[tex]\[ \begin{cases} x_1 + 2 x_2 = 120 \\ 2 x_1 + x_2 = 140 \end{cases} \][/tex]
Multiplying the first equation by 2:
[tex]\[ 2 x_1 + 4 x_2 = 240 \\ 2 x_1 + x_2 = 140 \][/tex]
Subtracting the second from the first:
[tex]\[ (2 x_1 + 4 x_2) - (2 x_1 + x_2) = 240 - 140 \\ 3 x_2 = 100 \\ x_2 = \frac{100}{3} \approx 33.33 \][/tex]
Substitute [tex]\( x_2 \)[/tex] into the second equation:
[tex]\[ 2 x_1 + \frac{100}{3} = 140 \\ 2 x_1 = 140 - \frac{100}{3} \\ 2 x_1 = \frac{320}{3} \\ x_1 = \frac{160}{3} \approx 53.33 \][/tex]
So, the intersection point is [tex]\( \left(\frac{160}{3}, \frac{100}{3}\right) \)[/tex].
### Step 3: Evaluate the Objective Function at Vertices
Calculate [tex]\( Z_{\max} \)[/tex] at the vertices:
1. At [tex]\( (0, 0) \)[/tex]:
[tex]\[ Z = 0 + \frac{3}{2} \times 0 = 0 \][/tex]
2. At [tex]\( (40, 40) \)[/tex]:
[tex]\[ Z = 40 + \frac{3}{2} \times 40 = 40 + 60 = 100 \][/tex]
3. At [tex]\( (60, 20) \)[/tex]:
[tex]\[ Z = 60 + \frac{3}{2} \times 20 = 60 + 30 = 90 \][/tex]
4. At [tex]\( \left(\frac{160}{3}, \frac{100}{3}\right) \)[/tex]:
[tex]\[ Z = \frac{160}{3} + \frac{3}{2} \times \frac{100}{3} = \frac{160}{3} + \frac{150}{3} = \frac{310}{3} \approx 103.33 \][/tex]
### Step 4: Optimal Solution and Determination of [tex]\( C_2 \)[/tex]
The current optimal solution is at [tex]\( \left(\frac{160}{3}, \frac{100}{3}\right) \)[/tex] with [tex]\( Z = 103.33 \)[/tex].
When the objective function changes to:
[tex]\[ Z_{\max } = x1 + C_2 x2 \][/tex]
We need this vertex to remain optimal:
[tex]\[ Z = \left(\frac{160}{3}\right) + C_2 \left(\frac{100}{3}\right) = 103.33 \][/tex]
Therefore:
[tex]\[ \frac{160}{3} + C_2 \left(\frac{100}{3}\right) \leq 103.33 \][/tex]
To determine the exact range:
1. Rewrite the relation:
[tex]\[ \frac{160 + 100 C_2 }{3} = 103.33 \][/tex]
[tex]\[ 160 + 100 C_2 = 310 \][/tex]
[tex]\[ 100 C_2 = 150 \][/tex]
[tex]\[ C_2 = 1.5 \][/tex]
Thus, this indicates that for [tex]\( \left(\frac{160}{3}, \frac{100}{3}\right) \)[/tex] to remain optimal, [tex]\( C_2 \leq 1.5 \)[/tex].
So, the present solution will be optimal within the range of [tex]\( C_2 \leq 1.5 \)[/tex].
