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Determine the optimal daily production mix given the following parameters:

Minimize [tex]\( z = 4x_1 + x_2 \)[/tex]

Subject to:
[tex]\[
\begin{aligned}
3x_1 + x_2 & = 30 \\
4x_1 + 3x_2 & \geq 60 \\
x_1 + 2x_2 & \leq 40 \\
x_1, x_2 & \geq 0
\end{aligned}
\][/tex]

The starting solution consists of artificial variables [tex]\( x_4 \)[/tex] and [tex]\( x_5 \)[/tex] for the first and second constraints, and [tex]\( x_6 \)[/tex] for the third constraint. Use [tex]\( M = 100 \)[/tex] for the artificial variables.


Sagot :

Certainly! To determine the optimal daily production mix given the constraints and the objective function, we'll follow these steps.

Objective:
Minimize [tex]\( z = 4x_1 + x_2 \)[/tex]

Subject to constraints:
1. [tex]\( 3x_1 + x_2 \leq 30 \)[/tex]
2. [tex]\( 4x_1 + 3x_2 \geq 60 \)[/tex]
3. [tex]\( x_1 + 2x_2 \leq 40 \)[/tex]
4. [tex]\( x_1, x_2 \geq 0 \)[/tex]

### Step-by-Step Solution:

1. Convert all inequalities to standard form:

To handle different forms (≥ or ≤), we transform them to standard "≤" form by multiplying inequalities with a negative sign if necessary.

- The first constraint already is a "≤" type:
[tex]\[ 3x_1 + x_2 \leq 30 \][/tex]

- The seconda constraint can be converted to a "≤":
[tex]\[ 4x_1 + 3x_2 \geq 60 \quad \text{which becomes} \quad -4x_1 - 3x_2 \leq -60 \][/tex]

- The third constraint is of "≤" type:
[tex]\[ x_1 + 2x_2 \leq 40 \][/tex]

2. Bounds:

We need to add the non-negative constraints for the variables:
[tex]\[ x_1 \geq 0, \quad x_2 \geq 0 \][/tex]

3. Set up the Linear Programming (LP) problem:

Now, we have the LP problem as:
[tex]\[ \begin{aligned} \text{Minimize} \quad & z = 4x_1 + x_2 \\ \text{subject to} \quad & 3x_1 + x_2 \leq 30, \\ & -4x_1 - 3x_2 \leq -60, \\ & x_1 + 2x_2 \leq 40, \\ & x_1 \geq 0, \\ & x_2 \geq 0. \end{aligned} \][/tex]

4. Solving the LP problem:

The optimal solution to this problem is [tex]\( x_1 = 0 \)[/tex] and [tex]\( x_2 = 20 \)[/tex].

The value of the objective function at this point is:
[tex]\[ z = 4(0) + 1(20) = 20 \][/tex]

### Result:

The optimal daily production mix is:
- [tex]\( x_1 = 0 \)[/tex]
- [tex]\( x_2 = 20 \)[/tex]

This gives us a minimized cost of [tex]\( z = 20 \)[/tex].

So, the minimum value of the objective function [tex]\( z \)[/tex] is 20, achieved when [tex]\( x_1 = 0 \)[/tex] and [tex]\( x_2 = 20 \)[/tex].