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

The correct answer is B: 4.5
Let's break down the problem step by step: 
Given information: (a+2b=6) 
We want to find the average of the maximum and minimum possible values of (a+b). 
Step 1: Expressing (a) in terms of (b) 
From (a+2b=6),we can solve for (a): 
[a=6-2b] 
Step 2: Finding Minimum and Maximum Values of (a+b) 
Substitute the expression for (a) into (a+b): 
[a+b=(6-2b)+b=6-b] 
Since (a) and (b) are non-negative, the minimum value of (a+b) occurs when (b) is maximum [which is (b=3)],and the maximum value of (a+b) occurs when (b) is minimum [which is (b=0)]. 
Minimum value of (a+b):(6-3=3) 
Maximum value of (a+b):(6-0=6) 
Step 3: Finding Average 
Average of maximum and minimum values:\(\frac{3 + 6}{2}=\frac{9}{2}=4.5\) 
Therefore,the average of the maximum and minimum possible values of (a+b) is 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): \(4.5\)