A commercial bakery starts a 30-day month with a 400-kilogram supply of a specialty flour. For the first 10 days...

1. TRANSLATE the problem information

• Given information:
  • Initial flour supply: 400 kg
  • Days 1-10: 15 kg used per day
  • Days 11-30: 8 kg used per day
  • Need function \(\mathrm{S(d)}\) for remaining flour where \(\mathrm{10 \leq d \leq 30}\)

2. INFER the solution strategy

• Since our function domain starts at \(\mathrm{d = 10}\), we need to:
  • First calculate how much flour remains at day 10
  • Then model the additional usage from day 10 onwards

3. Calculate flour remaining at day 10

• Flour used in first 10 days: \(\mathrm{10 \times 15 = 150}\) kg
• Flour remaining at day 10: \(\mathrm{400 - 150 = 250}\) kg

4. INFER the usage pattern for \(\mathrm{d \geq 10}\)

• When we're at day \(\mathrm{d}\) (where \(\mathrm{d \geq 10}\)), we've had \(\mathrm{(d - 10)}\) additional days since day 10
• In those \(\mathrm{(d - 10)}\) days, we used 8 kg per day
• Total additional flour used: \(\mathrm{8(d - 10)}\) kg

5. TRANSLATE this into a function

• Remaining flour = (Flour at day 10) - (Additional flour used since then)
• \(\mathrm{S(d) = 250 - 8(d - 10)}\)

6. SIMPLIFY to match answer format

\(\mathrm{S(d) = 250 - 8(d - 10)}\)
\(\mathrm{S(d) = 250 - 8d + 80}\)
\(\mathrm{S(d) = 330 - 8d}\)

Answer: (B) \(\mathrm{S(d) = 330 - 8d}\)

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak INFER skill: Students don't recognize that the function domain starting at \(\mathrm{d = 10}\) means they need to calculate from day 10 forward, not from day 1.

Instead, they might try to create a single function covering all 30 days, leading to confusion about how to handle the rate change. This often causes them to incorrectly use the 8 kg/day rate for all days, giving them \(\mathrm{S(d) = 400 - 8d}\), leading them to select Choice (C) (\(\mathrm{S(d) = 400 - 8d}\)).

Second Most Common Error:

Poor SIMPLIFY execution: Students correctly set up \(\mathrm{S(d) = 250 - 8(d - 10)}\) but make algebraic errors when expanding.

Common mistake: \(\mathrm{250 - 8(d - 10) = 250 - 8d - 80 = 170 - 8d}\) (forgetting that \(\mathrm{-8 \times (-10) = +80}\)). This doesn't match any answer choice exactly, leading to confusion and guessing.

The Bottom Line:

This problem requires students to understand that piecewise functions often need reference points other than the origin, and that careful algebra is essential when converting between equivalent forms of linear expressions.