In an apartment complex, the number of people aged $51$ years and above is $30$ and there are at most $39$ people whose ages are below $51$ years. The average age of all the people in the apartment complex is $38$ years. What is the largest possible average age, in years, of the people whose ages are below $51$ years?

We have an apartment complex with two age groups:

30 people aged 51 years and above

At most 39 people aged below 51 years

Overall average age is 38 years

We need to find the largest possible average age of people below 51 years.

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Let's define our variables:

A₁ = average age of people aged 51 and above

A₂ = average age of people aged below 51 years (this is what we want to maximize)

N₂ = number of people aged below 51 years

Key insight: We'll use the weighted average formula since we have two groups with different sizes.

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The overall average of 38 years comes from combining both groups:

$\text{Overall Average} = \tfrac{\text{(Average₁ × Count₁) + (Average₂ × Count₂)}}{\text{Total Count}}$

Substituting our values:

$38 = \tfrac{A_1 \times 30 + A_2 \times N_2}{30 + N_2}$

Why this formula works: Each group contributes to the total based on both their average age and how many people are in that group.

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$38(30 + N_2) = A_1 \times 30 + A_2 \times N_2$

$1140 + 38N_2 = 30A_1 + N_2A_2$

$1140 = 30A_1 + N_2(A_2 - 38)$

What this tells us: The sum of weighted deviations from 38 must equal 1140.

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From our equation: $1140 = 30A_1 + N_2(A_2 - 38)$

To maximize A₂, we need to:

Minimize A₁: Since people in this group are "51 years and above," the minimum possible value is A₁ = 51

Maximize N₂: We're told "at most 39 people" below 51, so maximum N₂ = 39

Why this works: Making A₁ smaller and N₂ larger gives us more "room" to increase A₂.

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Substituting A₁ = 51 and N₂ = 39:

$1140 = 30(51) + 39(A_2 - 38)$

$1140 = 1530 + 39(A_2 - 38)$

$39(A_2 - 38) = -390$

$A_2 - 38 = -10$

$A_2 = 28$

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Therefore, the largest possible average age of people below 51 years is 28 years.

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When maximizing one variable in a weighted average problem, we look for ways to:

Minimize other variables that compete for the same "total"

Maximize the weight (count) of the group we're trying to optimize

This approach works because weighted averages create trade-offs between different groups.