If $a$ and $b$ are non-negative real numbers such that $a + 2b = 6$, then the average of the maximum and minimum possible values of $(a + b)$ is

We're given that $a + 2b = 6$ where $a, b \geq 0$ (non-negative real numbers).

Key insight: We need to express $(a+b)$ in terms of just one variable to find its maximum and minimum values.

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From the constraint $a + 2b = 6$, we can solve for $a$:

$a = 6 - 2b$

Now we can express $(a+b)$ in terms of $b$ only:

$a + b = (6 - 2b) + b = 6 - b$

Why this helps: Now we can see that $(a+b)$ depends only on $b$, making it easier to find extremes.

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Since both $a$ and $b$ must be non-negative:

For $a \geq 0$:

$6 - 2b \geq 0$

$6 \geq 2b$

$b \leq 3$

For $b \geq 0$:

$b \geq 0$

Therefore: $0 \leq b \leq 3$

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Since $(a + b) = 6 - b$:

Maximum value occurs when $b$ is at its minimum value:

When $b = 0$: $a + b = 6 - 0 = 6$

Check: $a = 6 - 2(0) = 6$

Minimum value occurs when $b$ is at its maximum value:

When $b = 3$: $a + b = 6 - 3 = 3$

Check: $a = 6 - 2(3) = 0$

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Average of maximum and minimum values:

$\text{Average} = \tfrac{\text{Maximum} + \text{Minimum}}{2} = \tfrac{6 + 3}{2} = \tfrac{9}{2} = 4.5$

The average of the maximum and minimum possible values of $(a+b)$ is 4.5.

Key Takeaway: When finding extremes of an expression with constraints, express the target expression in terms of one variable, then find the valid range for that variable. The extremes occur at the endpoints of this range.