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].
