Given: \(a + 2b = 6\)
Determine the average of the highest and lowest possible values of \(a + b\).
Step 1: Express \( a \) in terms of \( b \)
From the given equation: \(a = 6 - 2b\)
Step 2: Substitute into \( a + b \)
\[a + b = (6 - 2b) + b = 6 - b\]Given the constraints \( a \geq 0 \) and \( b \geq 0 \) (non-negative), we establish the bounds for \( b \):
- \( a = 6 - 2b \geq 0 \Rightarrow b \leq 3 \)
- \( b \geq 0 \)
Therefore, the range for \( b \) is: \(0 \leq b \leq 3\)
Step 3: Determine the maximum and minimum values of \( a + b \)
- When \(b = 0\), the value of \(a + b\) is \( 6 - 0 = 6 \) (Maximum).
- When \(b = 3\), the value of \(a + b\) is \( 6 - 3 = 3 \) (Minimum).
Step 4: Calculate the average
\[\text{Average} = \frac{\text{Maximum Value} + \text{Minimum Value}}{2} = \frac{6 + 3}{2} = \frac{9}{2} = 4.5\]Final Answer: \(\boxed{4.5}\)