Question:medium

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

Updated On: Jan 15, 2026

  • 4

  • 4.5

  • 3.5

  • 3

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The Correct Option is B

Solution and Explanation

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: \(\boxed{4.5}\)

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