Question

Consider the Linear Programming Problem:
Minimize 3x₁+4x₂ +2x₃
subject to
x₁+x₂+x₃≤6
x₁+2x₂+x₃≤10
x₁,x₂,x₃≥0.
Then, the number of basic solutions are

• A. 7
• B. 9
• C. 10
• D. 8
• E. 3

Answer 1

The Correct Option is C

Given the linear programming problem:

\[ \begin{align*} \text{Minimize } & 3x_1 + 4x_2 + 2x_3 \\ \text{subject to} & \\ & x_1 + x_2 + x_3 \leq 6 \\ & x_1 + 2x_2 + x_3 \leq 10 \\ & x_1, x_2, x_3 \geq 0 \end{align*} \]

To find the number of basic solutions:

Step 1: Convert inequalities to equations by adding slack variables \( s_1, s_2 \geq 0 \): \[ \begin{align*} x_1 + x_2 + x_3 + s_1 &= 6 \\ x_1 + 2x_2 + x_3 + s_2 &= 10 \end{align*} \]

Step 2: The system now has 5 variables (\( x_1, x_2, x_3, s_1, s_2 \)) and 2 equations.

Step 3: The number of basic solutions is equal to the number of ways to choose 2 basic variables from 5, which is: \[ \binom{5}{2} = 10 \]

Step 4: However, we must consider only feasible solutions (non-negative values). All 10 combinations need to be checked for feasibility, but the total possible basic solutions is 10.

The correct answer is (C) 10.

Answer 2

To find the number of basic solutions, we need to examine the constraints of the linear programming problem. We have three variables (\(x_1, x_2, x_3\)) and two inequality constraints. We can convert the inequalities into equalities by introducing slack variables:

\[ x_1 + x_2 + x_3 + s_1 = 6 \] \[ x_1 + 2x_2 + x_3 + s_2 = 10 \] \[ x_1, x_2, x_3, s_1, s_2 \geq 0 \]

Now we have a system of two equations with five variables. A basic solution is obtained by setting three variables to zero and solving for the remaining two. The number of ways to choose three variables out of five to set to zero is given by the combination formula:

\[ \binom{5}{3} = \binom{5}{2} = \frac{5!}{3!2!} = \frac{5 \cdot 4}{2 \cdot 1} = 10 \]

However, we must ensure that the remaining two variables are non-negative. Let's analyze this systematically:

Basic Solutions:

There are 10 combinations of setting three variables to zero. Let's examine each one to see if a feasible solution exists.

1. \(x_1 = x_2 = x_3 = 0\): \(s_1 = 6, s_2 = 10\) (Feasible)
2. \(x_1 = x_2 = s_1 = 0\): \(x_3 = 6, s_2 = 4\) (Feasible)
3. \(x_1 = x_2 = s_2 = 0\): \(x_3 = 10, s_1 = -4\) (Infeasible, \(s_1\) must be non-negative)
4. \(x_1 = x_3 = s_1 = 0\): \(x_2 = 6, s_2 = -2\) (Infeasible)
5. \(x_1 = x_3 = s_2 = 0\): \(x_2 = 5, s_1 = 1\) (Feasible)
6. \(x_2 = x_3 = s_1 = 0\): \(x_1 = 6, s_2 = 4\) (Feasible)
7. \(x_2 = x_3 = s_2 = 0\): \(x_1 = 10, s_1 = -4\) (Infeasible)
8. \(x_1 = s_1 = s_2 = 0\): \(x_2 = 5, x_3 = 1\) (Feasible)
9. \(x_2 = s_1 = s_2 = 0\): \(x_1 = 6, x_3 = 4\) (Feasible)
10. \(x_3 = s_1 = s_2 = 0\): \(x_1 = 6, x_2 = 4\) (Feasible)

Thus, there are 10 basic solutions.