Sure, let's solve this linear programming problem step-by-step:
### Problem Statement
We are given the following linear programming problem:
- Objective: Minimize the cost [tex]\(Z = 0.20x + 0.30y\)[/tex]
- Subject to constraints:
[tex]\[
\begin{align*}
3x + 5y & \geq 45 \\
2x + y & \geq 20 \\
x, y & \geq 0
\end{align*}
\][/tex]
### Step-by-Step Solution
1. Graph the Constraints:
We need to convert the inequalities into equations for graphing purposes.
- [tex]\(3x + 5y = 45\)[/tex]
- [tex]\(2x + y = 20\)[/tex]
2. Plot the Constraints:
- For [tex]\(3x + 5y = 45\)[/tex]:
[tex]\[
\begin{cases}
x = 0 \implies y = 9 \quad \text{(0, 9)} \\
y = 0 \implies x = 15 \quad \text{(15, 0)}
\end{cases}
\][/tex]
Thus, the line intersects the y-axis at (0, 9) and the x-axis at (15, 0).
- For [tex]\(2x + y = 20\)[/tex]:
[tex]\[
\begin{cases}
x = 0 \implies y = 20 \quad \text{(0, 20)} \\
y = 0 \implies x = 10 \quad \text{(10, 0)}
\end{cases}
\][/tex]
Thus, the line intersects the y-axis at (0, 20) and the x-axis at (10, 0).
3. Determine the Feasible Region:
The feasible region is the area where both constraints overlap, and [tex]\(x, y \geq 0\)[/tex].
4. Find the Intersection Points:
- Solve [tex]\(3x + 5y = 45\)[/tex] and [tex]\(2x + y = 20\)[/tex]:
Multiply [tex]\(2x + y = 20\)[/tex] by 5:
[tex]\[
10x + 5y = 100
\][/tex]
Subtract [tex]\(3x + 5y = 45\)[/tex] from [tex]\(10x + 5y = 100\)[/tex]:
[tex]\[
7x = 55 \implies x = \frac{55}{7} \implies x \approx 7.857
\][/tex]
Substitute [tex]\(x\)[/tex] into [tex]\(2x + y = 20\)[/tex]:
[tex]\[
2 \left(\frac{55}{7}\right) + y = 20 \implies y = 20 - \frac{110}{7} \implies y = \frac{140}{7} - \frac{110}{7} \implies y = \frac{30}{7} \implies y \approx 4.286
\][/tex]
So the intersection point (vertex of feasible region) is approximately [tex]\((7.857, 4.286)\)[/tex].
5. Evaluate the Objective Function:
We substitute the critical points (the vertices of the feasible region) into the objective function [tex]\(Z = 0.20x + 0.30y\)[/tex].
At point [tex]\((7.857, 4.286)\)[/tex]:
[tex]\[
Z = 0.20 \times 7.857 + 0.30 \times 4.286 \approx 1.571 + 1.286 = 2.857
\][/tex]
6. Conclusion:
The minimum cost occurs at [tex]\(x \approx 7.857\)[/tex] and [tex]\(y \approx 4.286\)[/tex], and the minimum cost is approximately [tex]\(Rs. 2.857\)[/tex].
Therefore, the minimum cost [tex]\(Z_{\min}\)[/tex] is approximately [tex]\(Rs. 2.857\)[/tex].
