Min Z = 5x + 7y
S.T.
x + y = 10
x ≥ 2
y ≥ 3
X ≥ 0
y ≥ 0

To solve the linear programming problem:

**Minimize**:  
\[ Z = 5x + 7y \]

**Subject to**:  
\[
\begin{aligned}
x + y &= 10 \\
x &\geq 2 \\
y &\geq 3 \\
x &\geq 0 \\
y &\geq 0
\end{aligned}
\]

### Steps to solve:

1. **Identify the constraints**:
   - The first constraint is \( x + y = 10 \), which represents a straight line in the xy-plane.
   - The second constraint \( x \geq 2 \) sets a minimum value for \( x \).
   - The third constraint \( y \geq 3 \) sets a minimum value for \( y \).
   - The non-negativity constraints \( x \geq 0 \) and \( y \geq 0 \) ensure that the solution lies in the first quadrant.

2. **Graph the constraints**:
   - \( x + y = 10 \) is a straight line that passes through the points (0, 10) and (10, 0).
   - \( x \geq 2 \) is a vertical line at \( x = 2 \), so only points to the right of this line are feasible.
   - \( y \geq 3 \) is a horizontal line at \( y = 3 \), so only points above this line are feasible.
   - The feasible region is bounded by these constraints.

3. **Find the feasible region**:
   The intersection of the lines forms a polygon, and the feasible region is where all the inequalities are satisfied:
   - \( x + y = 10 \)
   - \( x \geq 2 \)
   - \( y \geq 3 \)

4. **Determine the corner points**:
   The corner points of the feasible region occur where the lines intersect:
   - When \( x = 2 \), solve for \( y \): \( 2 + y = 10 \Rightarrow y = 8 \). So one point is \( (2, 8) \).
   - When \( y = 3 \), solve for \( x \): \( x + 3 = 10 \Rightarrow x = 7 \). So another point is \( (7, 3) \).
   - A third point occurs at the intersection of \( x = 2 \) and \( y = 3 \), which gives the point \( (2, 3) \).

5. **Evaluate Z at each corner point**:
   - At \( (2, 8) \): \( Z = 5(2) + 7(8) = 10 + 56 = 66 \)
   - At \( (7, 3) \): \( Z = 5(7) + 7(3) = 35 + 21 = 56 \)
   - At \( (2, 3) \): \( Z = 5(2) + 7(3) = 10 + 21 = 31 \)

6. **Conclusion**:
   The minimum value of \( Z \) is 31, which occurs at the point \( (2, 3) \).

### Solution:
The optimal solution is \( x = 2 \), \( y = 3 \), and the minimum value of \( Z \) is **31**.

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Would you like more details on any step? Here are some related questions:

1. How can we graph the feasible region for this problem?
2. What are the typical steps in solving linear programming problems?
3. Can this method be applied to maximize an objective function?
4. How do the constraints affect the shape of the feasible region?
5. How can the Simplex method be used for linear programming?

**Tip**: In linear programming, the minimum or maximum value will always occur at a corner point of the feasible region.

Min Z = 5x + 7y
S.T.
x + y = 10
x ≥ 2
y ≥ 3
x ≥ 0
y ≥ 0

This problem demonstrates how to solve a linear programming optimization by minimizing Z = 5x + 7y, subject to constraints x + y = 10, x ≥ 2, y ≥ 3, x ≥ 0, and y ≥ 0. The solution involves identifying the feasible region and calculating the objective function at the corner points. Suitable for students in Grades 10-12.

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