Problem Analysis

We need to find the maximum value of the objective function $f(X, Y) = 3X + 6Y$ given the constraints:

• $2X + Y \le 6$
• $X + 2Y \le 8$
• $X > 0$
• $Y > 0$

This is a linear programming problem. The maximum value of a linear function subject to linear constraints occurs at one of the vertices (corner points) of the feasible region defined by the constraints.

Finding the Critical Vertex

The critical vertex within the feasible region is often found at the intersection of the boundary lines of the main constraints. We solve the system of equations:

1. $2X + Y = 6$
2. $X + 2Y = 8$

From equation (1), we get $Y = 6 - 2X$. Substituting this into equation (2):

$X + 2(6 - 2X) = 8$

$X + 12 - 4X = 8$

Combining terms:

$-3X = 8 - 12$

$-3X = -4$

$X = \frac{4}{3}$

Now, substitute the value of $X$ back into the expression for $Y$:

$Y = 6 - 2\left(\frac{4}{3}\right) = 6 - \frac{8}{3} = \frac{18}{3} - \frac{8}{3} = \frac{10}{3}$

The intersection point is $\left(\frac{4}{3}, \frac{10}{3}\right)$.

Verifying Options and Evaluating the Function

We check if the given options satisfy the constraints and evaluate the objective function $f(X, Y) = 3X + 6Y$ at the valid points.

Option 1: $\left(\frac{4}{3}, \frac{10}{3}\right)$

• Check constraints:
  • $2\left(\frac{4}{3}\right) + \frac{10}{3} = \frac{8}{3} + \frac{10}{3} = \frac{18}{3} = 6 \le 6$ (Satisfied)
  • $\frac{4}{3} + 2\left(\frac{10}{3}\right) = \frac{4}{3} + \frac{20}{3} = \frac{24}{3} = 8 \le 8$ (Satisfied)
  • $X = \frac{4}{3} > 0$ and $Y = \frac{10}{3} > 0$ (Satisfied)
• Evaluate $f(X, Y)$: $f\left(\frac{4}{3}, \frac{10}{3}\right) = 3\left(\frac{4}{3}\right) + 6\left(\frac{10}{3}\right) = 4 + 2(10) = 4 + 20 = 24$

Option 2: $\left(\frac{8}{3}, \frac{20}{3}\right)$

• Check constraints:
  • $2\left(\frac{8}{3}\right) + \frac{20}{3} = \frac{16}{3} + \frac{20}{3} = \frac{36}{3} = 12$. Since $12 \not\le 6$, this point is not feasible.

Option 3: $\left(\frac{8}{3}, \frac{10}{3}\right)$

• Check constraints:
  • $2\left(\frac{8}{3}\right) + \frac{10}{3} = \frac{16}{3} + \frac{10}{3} = \frac{26}{3}$. Since $\frac{26}{3} \approx 8.67 \not\le 6$, this point is not feasible.

Option 4: $\left(\frac{4}{3}, \frac{20}{3}\right)$

• Check constraints:
  • $2\left(\frac{4}{3}\right) + \frac{20}{3} = \frac{8}{3} + \frac{20}{3} = \frac{28}{3}$. Since $\frac{28}{3} \approx 9.33 \not\le 6$, this point is not feasible.

Conclusion

Only Option 1, $\left(\frac{4}{3}, \frac{10}{3}\right)$, lies within the feasible region defined by the constraints. Therefore, it yields the maximum value for the function $f(X, Y)$. The maximum value is 24.