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We are asked to find the minimum value of \( C = 2x + y \) under the given set of constraints:

\[
\begin{aligned}
1. & \quad 2x + y \leq 10 \\
2. & \quad x - 3y \geq -3 \\
3. & \quad x \geq 0 \\
4. & \quad y \geq 2
\end{aligned}
\]

This is a linear programming problem, and we can solve it by identifying the vertices of the feasible region and evaluating the objective function \( C = 2x + y \) at those points. The solution steps are:

### 1. Convert inequalities to equations:
   - \( 2x + y = 10 \)
   - \( x - 3y = -3 \)
   - \( x = 0 \)
   - \( y = 2 \)

### 2. Find the vertices by solving the system of equations formed by pairs of these lines.

#### Intersection of \( 2x + y = 10 \) and \( x - 3y = -3 \):
Solve this system by substitution or elimination:
- From \( x - 3y = -3 \), we have \( x = 3y - 3 \).
- Substitute into \( 2x + y = 10 \):
  \[
  2(3y - 3) + y = 10 \\
  6y - 6 + y = 10 \\
  7y = 16 \\
  y = \frac{16}{7}
  \]
  Now, substitute \( y = \frac{16}{7} \) into \( x = 3y - 3 \):
  \[
  x = 3\left(\frac{16}{7}\right) - 3 = \frac{48}{7} - 3 = \frac{27}{7}
  \]
  So, one vertex is \( \left( \frac{27}{7}, \frac{16}{7} \right) \).

#### Intersection of \( 2x + y = 10 \) and \( y = 2 \):
Substitute \( y = 2 \) into \( 2x + y = 10 \):
\[
2x + 2 = 10 \\
2x = 8 \\
x = 4
\]
So, the second vertex is \( (4, 2) \).

#### Intersection of \( x - 3y = -3 \) and \( y = 2 \):
Substitute \( y = 2 \) into \( x - 3y = -3 \):
\[
x - 3(2) = -3 \\
x - 6 = -3 \\
x = 3
\]
So, the third vertex is \( (3, 2) \).

#### Intersection of \( x = 0 \) and \( x - 3y = -3 \):
Substitute \( x = 0 \) into \( x - 3y = -3 \):
\[
0 - 3y = -3 \\
y = 1
\]
However, this point \( (0, 1) \) does not satisfy the constraint \( y \geq 2 \), so it is not valid.

#### Intersection of \( x = 0 \) and \( 2x + y = 10 \):
Substitute \( x = 0 \) into \( 2x + y = 10 \):
\[
y = 10
\]
So, the fourth vertex is \( (0, 10) \).

### 3. Evaluate \( C = 2x + y \) at each vertex:
- At \( \left( \frac{27}{7}, \frac{16}{7} \right) \): 
  \[
  C = 2\left(\frac{27}{7}\right) + \frac{16}{7} = \frac{54}{7} + \frac{16}{7} = \frac{70}{7} = 10
  \]
- At \( (4, 2) \): 
  \[
  C = 2(4) + 2 = 8 + 2 = 10
  \]
- At \( (3, 2) \): 
  \[
  C = 2(3) + 2 = 6 + 2 = 8
  \]
- At \( (0, 10) \): 
  \[
  C = 2(0) + 10 = 10
  \]

### 4. Conclusion:
The minimum value of \( C = 8 \) occurs at the point \( (3, 2) \).

Would you like further details or have any questions? 

Here are 5 related questions:
1. How do we graph the feasible region for this problem?
2. Can linear programming problems have multiple solutions?
3. What if one of the constraints was removed—how would the solution change?
4. How is the Simplex method used for larger linear programming problems?
5. Why do we evaluate the objective function at vertices in linear programming?

**Tip**: When solving linear programming problems, always check the feasibility of points with respect to all constraints before determining the optimal value.

Find the minimum value of C = 2x + y subject to the constraints: 2x + y ≤ 10, x - 3y ≥ -3, x ≥ 0, y ≥ 2.

Learn how to find the minimum value of the objective function C = 2x + y given multiple linear constraints, including inequalities. This linear programming problem is suitable for students in Grades 10-12 and involves solving inequalities and evaluating vertices.

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