Question

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

• A. 4
• B. 4.5
• C. 3.5
• D. 3

Answer 1

The Correct Option is B

Given: \(a + 2b = 6\)

We are asked to find the average of the maximum and minimum possible values of \(a + b\).

Step 1: Express \( a \) in terms of \( b \)

From the equation: \(a = 6 - 2b\)

Step 2: Substitute into \( a + b \)

\[a + b = (6 - 2b) + b = 6 - b\]

Now, since \( a \geq 0 \) and \( b \geq 0 \) (non-negative), we find limits on \( b \):

• \( a = 6 - 2b \geq 0 \Rightarrow b \leq 3 \)
• \( b \geq 0 \)

So, \(0 \leq b \leq 3\)

Step 3: Find maximum and minimum values of \( a + b \)

• When \(b = 0\), then \(a + b = 6 - 0 = 6\) (Maximum)
• When \(b = 3\), then \(a + b = 6 - 3 = 3\) (Minimum)

Step 4: Compute the average

\[\text{Average} = \frac{\text{Maximum} + \text{Minimum}}{2} = \frac{6 + 3}{2} = \frac{9}{2} = 4.5\]

Final Answer: \(\boxed{4.5}\)

Answer 2

Given: \(a + 2b = 6\)
From the above equation, we can say that maximum value b can take is \(3\) and minimum value b can take is \(0\).
\(a + b + b = 6\)
\(a + b = 6 - b\)
\(a + b\) is maximum when \(b\) is minimum.
\(⇒\) \(b = 0\)
Maximum value of \(a + b\)
\(= 6 - 0\)
\(= 6\)
\(a + b\) is minimum when b is maximum,
\(⇒ b = 3\)
Minimum value of \(a + b\)
\(= 6 - 3\)
\(= 3 \)
Average \(=\frac {6+3}{2}\)
\(=\frac 92\)
= 4.5

So, the correct option is (B): \(4.5\)