Question

Let the linear programming problem be
Maximize Z = - 0.2x₁ + x₂
subject to 2x₁ + 5x₂ ≤ 70,
x₁ + x₂ ≤ 20,
x₁, x₂ ≥ 0.
If x₁ = a and x₂ = b is the optimal solution, then a+b=______ (in integer).

Answer 1

Correct Answer: 14

To solve the given linear programming problem, we first determine the feasible region using the constraints. The constraints are:
1. \( 2x_1 + 5x_2 \leq 70 \)
2. \( x_1 + x_2 \leq 20 \)
3. \( x_1 \geq 0 \), \( x_2 \geq 0 \)

The objective function to maximize is \( Z = -0.2x_1 + x_2 \). We find the intersection points of the constraints to identify the corner points of the feasible region.

**Step 1: Find intersection points of constraints**
**Constraint 1 and Constraint 2:**
Solving \( 2x_1 + 5x_2 = 70 \) and \( x_1 + x_2 = 20 \) simultaneously:
From the second equation: \( x_1 = 20 - x_2 \)
Substitute into the first equation: \( 2(20 - x_2) + 5x_2 = 70 \) 
\( 40 - 2x_2 + 5x_2 = 70 \) 
\( 3x_2 = 30 \) 
\( x_2 = 10 \) 
Substitute \( x_2 = 10 \) back into \( x_1 = 20 - x_2 \):
\( x_1 = 10 \)
Intersection point: \( (10, 10) \)

**Intersection with axes:**
For \( 2x_1 + 5x_2 = 70 \):
When \( x_1 = 0 \), \( 5x_2 = 70 \), \( x_2 = 14 \)
When \( x_2 = 0 \), \( 2x_1 = 70 \), \( x_1 = 35 \) (not within the constraint \( x_1 + x_2 \leq 20 \))
Feasible intersection: \( (0, 14) \)

For \( x_1 + x_2 = 20 \):
When \( x_1 = 0 \), \( x_2 = 20 \) (not feasible as it violates \( 2x_1 + 5x_2 \leq 70 \))
When \( x_2 = 0 \), \( x_1 = 20 \) (not feasible as it violates \( 2x_1 + 5x_2 \leq 70 \))
Feasible point: \( (10, 10) \)

**Step 2: Evaluate the objective function at feasible region's corner points**
1. At \( (0, 0) \):
\( Z = -0.2(0) + (0) = 0 \)
2. At \( (0, 14) \):
\( Z = -0.2(0) + 14 = 14 \)
3. At \( (10, 10) \):
\( Z = -0.2(10) + 10 = 8 \) (After calculation)

The maximum value of \( Z \) is 14, which occurs at \( (0, 14) \). Therefore, \( a = 0 \), \( b = 14 \), resulting in \( a+b = 0 + 14 = 14 \).

Hence, the solution \( a+b = 14 \) is correct.