Three boxes of supplies have an average (arithmetic mean) weight of 7text{ kilograms} and a median weight of 9text{ kilograms}....

1. Translate the problem requirements: We have three boxes with average weight \(7\) kg and median weight \(9\) kg. We need to find the maximum possible weight of the lightest box (the smallest of the three weights).
2. Use median constraint to organize the boxes: Since median is \(9\) kg, the middle box weighs \(9\) kg, so we can label the boxes as: lightest \(\leq 9 \leq\) heaviest.
3. Apply the average constraint: The sum of all three weights must equal \(3 \times 7 = 21\) kg, which gives us lightest \(+ 9 +\) heaviest \(= 21\).
4. Maximize the lightest box weight: To maximize the lightest box, we need to minimize the heaviest box while respecting the constraint that heaviest \(\geq 9\).

Execution of Strategic Approach

1. Translate the problem requirements

Let's start by understanding what we know and what we're looking for in plain English.

We have three boxes of supplies. Think of them sitting on a table in front of us. We know two key facts about their weights:

• The average weight is \(7\) kilograms
• The median weight is \(9\) kilograms

Our goal is to find the maximum possible weight of the lightest box. In other words, we want to make the smallest weight as large as possible while still satisfying our constraints.

Process Skill: TRANSLATE - Converting the problem language into clear mathematical understanding

2. Use median constraint to organize the boxes

Let's think about what the median tells us. When we have three numbers arranged in order from smallest to largest, the median is simply the middle number.

Since our median is \(9\) kg, this means the middle box weighs exactly \(9\) kg. So if we arrange our three boxes from lightest to heaviest, we get:

Lightest box \(\leq 9\) kg \(\leq\) Heaviest box

This is a crucial insight because it fixes the weight of our middle box at \(9\) kg, giving us a concrete starting point.

Mathematically, if we call the weights L (lightest), M (middle), and H (heaviest), then:
\(\mathrm{L} \leq \mathrm{M} = 9 \leq \mathrm{H}\)

3. Apply the average constraint

Now let's use the average constraint. When three boxes have an average weight of \(7\) kg, this means their total weight divided by \(3\) equals \(7\).

In everyday terms: if you put all three boxes on a scale, the total weight would be \(3 \times 7 = 21\) kg.

Since we know the middle box weighs \(9\) kg, we can write:
Lightest \(+ 9 +\) Heaviest \(= 21\)

Solving for the relationship between lightest and heaviest:
Lightest + Heaviest \(= 21 - 9 = 12\)

So: Heaviest \(= 12 -\) Lightest

Process Skill: APPLY CONSTRAINTS - Using both given conditions to create solvable relationships

4. Maximize the lightest box weight

Now for the key insight: to maximize the weight of the lightest box, we need to think about what limits it.

From our median constraint, we know that the heaviest box must weigh at least \(9\) kg (since it can't be lighter than the median). So:
Heaviest \(\geq 9\)

Substituting our relationship from step 3:
\(12 -\) Lightest \(\geq 9\)
\(12 - 9 \geq\) Lightest
\(3 \geq\) Lightest

This means the lightest box can weigh at most \(3\) kg.

Let's verify this works: If the lightest box weighs \(3\) kg, then the heaviest box weighs \(12 - 3 = 9\) kg.
Our three weights would be: \(3, 9, 9\)

• Average: \((3 + 9 + 9) \div 3 = 21 \div 3 = 7\) ✓
• Median: \(9\) ✓
• All constraints satisfied ✓

Process Skill: CONSIDER ALL CASES - Recognizing that maximizing one variable requires minimizing others within constraints

Final Answer

The maximum possible weight of the lightest box is \(3\) kilograms.

This occurs when the three boxes weigh \(3\) kg, \(9\) kg, and \(9\) kg respectively, which satisfies both the average constraint (\(7\) kg) and median constraint (\(9\) kg).

The answer is (C) \(3\).

Common Faltering Points

Errors while devising the approach

1. Misinterpreting what "maximum weight of the lightest box" means

Students often confuse this with finding the maximum weight among all three boxes, rather than understanding we want to make the smallest weight as large as possible while satisfying constraints.

2. Not properly utilizing the median constraint

Many students forget that with three values, the median fixes the middle value at exactly \(9\) kg. They might treat this as just "one of the boxes weighs \(9\) kg" without recognizing the ordering constraint (lightest \(\leq 9 \leq\) heaviest).

3. Setting up the wrong optimization direction

Students may try to maximize the lightest box without recognizing that this requires minimizing the heaviest box within the given constraints.

Errors while executing the approach

1. Arithmetic errors in constraint manipulation

When solving \(12 -\) Lightest \(\geq 9\), students commonly make algebraic mistakes, such as getting Lightest \(\geq 3\) instead of Lightest \(\leq 3\), or incorrectly calculating \(12 - 9\).

2. Incorrectly applying the median constraint

Students might set up the constraint as Heaviest \(> 9\) (strictly greater) instead of Heaviest \(\geq 9\), leading them to conclude the maximum lightest weight is less than \(3\).

3. Failing to verify the solution satisfies all constraints

Students may arrive at the correct answer but fail to check that weights \((3, 9, 9)\) actually give average \(= 7\) and median \(= 9\), missing computational errors.

Errors while selecting the answer

No likely faltering points

Once students correctly execute the approach and find that the lightest box can weigh at most \(3\) kg, the answer selection is straightforward since \(3\) directly corresponds to choice (C).