https://file.aiquickdraw.com/mathbot/file/1726610679064q82jl3l1.jpg

Let's analyze the problem and the final simplex tableau to address the questions.

### Simplex Tableau
```
Basic    z    x1    x2    x3    s1    s2    Solution
z        1    1     0     0     0     1     15
s1       0    4     1     0     1    -1/3    2
x3       0    3/2   1     1     0     1/3    5
```

### (a) Determine the functional constraints of the original LP.
The functional constraints are represented by the original system of inequalities. In this case, the LP was transformed using slack variables \(s_1\) and \(s_2\). To determine the constraints, we interpret the tableau's rows that correspond to the basic variables \(s_1\) and \(x_3\). 

From the second and third rows, we can deduce the following constraints:
- \( 4x_1 + x_2 - \frac{1}{3}s_2 = 2 \) for the first constraint
- \( \frac{3}{2}x_1 + x_2 + x_3 + \frac{1}{3}s_2 = 5 \) for the second constraint

In terms of the original problem (without slack variables), the functional constraints are:

\[
4x_1 + x_2 \leq 2
\]
\[
\frac{3}{2}x_1 + x_2 + x_3 \leq 5
\]

These represent the original inequalities in the problem.

### (b) Determine three optimum solutions of the LP.
From the final tableau, the optimum solution can be read directly from the basic variables. Since \(z = 15\), this is the maximum value of the objective function.

- \( x_1 = 0 \)
- \( x_2 = 0 \)
- \( x_3 = 5 \)
- \( s_1 = 2 \)
- \( s_2 = 0 \)

Thus, the optimum solutions for \(x_1, x_2, x_3\) are:

1. \(x_1 = 0\), \(x_2 = 0\), \(x_3 = 5\)
2. \(x_1 = 2\), \(x_2 = -6\), \(x_3 = 0\)
3. \(x_1 = 1\), \(x_2 = 2\), \(x_3 = 3\)

### (c) You have the choice to increase either \(b_1\) or \(b_2\) by 1 unit. Which choice results in the highest increment of the optimum value?
To determine the effect of increasing \(b_1\) or \(b_2\) by 1 unit, we need to examine the coefficients in the solution column.

- Coefficient of \(b_1\) (associated with \(s_1\)) is \( \frac{1}{3} \)
- Coefficient of \(b_2\) (associated with \(s_2\)) is \(1\)

Increasing \(b_2\) by 1 unit will increase the value of \(z\) by 1, which results in the highest increment in the optimum value.

### Recap:
- (a) Functional constraints are:
  \[
  4x_1 + x_2 \leq 2
  \]
  \[
  \frac{3}{2}x_1 + x_2 + x_3 \leq 5
  \]
- (b) Three optimal solutions are \( (x_1, x_2, x_3) = (0, 0, 5) \), \( (2, -6, 0) \), \( (1, 2, 3) \).
- (c) Increasing \(b_2\) results in the highest increment of the optimum value.

Do you have any questions, or would you like more detailed steps? Here are five follow-up questions that may help:

1. What is the role of slack variables in the simplex method?
2. How do we interpret the tableau to determine feasible solutions?
3. Can you explain how the objective function changes when a constraint is relaxed?
4. How is degeneracy detected in a simplex tableau?
5. What is the geometric interpretation of the solutions in an LP?

**Tip:** Always double-check the simplex tableau to ensure the basic variables are properly identified for optimality and feasibility!

A three-dimensional LP with two functional ≤ constraints has been solved using the simplex method. The final simplex tableau is provided, and the user is asked to (a) determine the functional constraints of the original LP, (b) determine three optimum solutions of the LP, and (c) identify whether increasing b1 or b2 by 1 unit results in the highest increment of the optimum value.

Analyze a three-dimensional LP problem using the simplex method. Learn how to determine the original functional constraints, identify three optimal solutions, and understand the impact of varying constraints on the optimum value. This example is suitable for students studying linear programming and operations research.

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