Min Z = 5x + 7y

S. T

 x + y = 10

 x ≥ 2

 y ≥ 3 

 x ≥ 0

 x ≥ 0 with graph

The graph above shows the constraints and the feasible region:

- The blue line represents the equation \( x + y = 10 \).
- The dashed red line marks the constraint \( x \geq 2 \).
- The dashed green line marks the constraint \( y \geq 3 \).
- The shaded region is the feasible area where all constraints are satisfied.

Now, to solve for the minimum value of \( Z = 5x + 7y \) within this region, we evaluate the objective function at the vertices of the feasible region. These vertices are the key points where constraints intersect.

Let me calculate the values of \( Z \) at these points.The values of \( Z = 5x + 7y \) at the vertices of the feasible region are:

- At \( (2, 8) \), \( Z = 66 \)
- At \( (2, 3) \), \( Z = 31 \)
- At \( (7, 3) \), \( Z = 56 \)

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

### Conclusion:
The minimum value of the objective function is \( Z = 31 \), and it occurs at \( x = 2 \), \( y = 3 \).

Would you like further details or have any other questions?

Here are some related questions you might want to explore:
1. What if the objective function had different coefficients?
2. How do we solve similar problems with multiple constraints?
3. Can we extend this problem to three variables?
4. How does changing a constraint affect the feasible region?
5. What methods are available to solve this problem algebraically?

**Tip:** In linear programming problems, always check the vertices of the feasible region to find the optimal solution.

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

Explore how to minimize Z = 5x + 7y using linear programming. The problem involves finding the feasible region defined by constraints x + y = 10, x ≥ 2, y ≥ 3, and solving for the minimum value of Z. Suitable for Grades 10-12, this solution includes graphical methods and evaluation of vertices.

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