Not sure how to get the maximum average from this. Any idea how to think about it?

Question

Consider six distinct natural numbers such that the average of the two smallest numbers is 14, and the average of the two largest numbers is 28. Then, the maximum possible value of the average of these six numbers is

D. Aarav · Accepted Answer

To get the maximum average of all six numbers, you need to fix the first and last pairs and then make the middle two as large as possible. The two smallest numbers average to 14, so their sum is 28. Similarly, the two largest average to 28, so their sum is 56.
Now you already have four out of six numbers with a fixed total of 28 + 56 = 84. To maximize the overall average, make the two middle numbers as large as possible but still less than the largest ones. That happens when the middle numbers are 27 and 29 (just before and after the largest pair). But 29 is already part of the largest pair, so take 25 and 26 as the middle two.
Then the six numbers are: a, b, 25, 26, 27, 29 where a + b = 28. So total sum = 28 + 25 + 26 + 27 + 29 = 135 Average = 135 / 6 = 22.5

V Ritik · Answer

This is a classic case of fixing ends and maximizing the middle. Here&rsquo;s how to break it down:

Average of two smallest numbers = 14 ,  sum = 28
Average of two largest numbers = 28 ,  sum = 56
You want to pick the highest possible values for the middle two numbers
Try middle numbers as 25 and 26, then arrange as: a, b, 25, 26, 27, 29
a + b = 28,  total sum = 135
Average = 135 &divide; 6 = 22.5

So, the best possible average is 22.5. You lock the ends, maximize what&rsquo;s left in the middle, and the rest follows smoothly.