A company models its monthly revenue R (in dollars) using the function \(\mathrm{R(p) = (p + 0.4p)(60 - p)}\), where...

1. TRANSLATE the problem information

• Given information:
  • Revenue function: \(\mathrm{R(p) = (p + 0.4p)(60 - p)}\)
  • Need to find: \(\mathrm{R(25)}\)

• What this tells us: We need to substitute \(\mathrm{p = 25}\) into the function and calculate the result

2. SIMPLIFY the function first

• Before substituting, combine like terms in the first factor:
  • \(\mathrm{p + 0.4p = 1p + 0.4p = 1.4p}\)

• So our function becomes: \(\mathrm{R(p) = (1.4p)(60 - p)}\)

3. TRANSLATE and substitute the given value

• Substitute \(\mathrm{p = 25}\) into \(\mathrm{R(p) = (1.4p)(60 - p)}\):
  • \(\mathrm{R(25) = (1.4 \times 25)(60 - 25)}\)

4. SIMPLIFY by calculating each factor

• Calculate the first factor: \(\mathrm{1.4 \times 25 = 35}\)
• Calculate the second factor: \(\mathrm{60 - 25 = 35}\)
• So we have: \(\mathrm{R(25) = 35 \times 35}\)

5. SIMPLIFY to find the final answer

• Calculate: \(\mathrm{35 \times 35 = 1,225}\)

Answer: D. 1,225

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak SIMPLIFY execution: Students don't combine like terms correctly at the beginning. They might leave the expression as \(\mathrm{(p + 0.4p)}\) and try to substitute directly, leading to confusing calculations like \(\mathrm{(25 + 0.4 \times 25)}\), which becomes \(\mathrm{(25 + 10)(35) = 35 \times 35 = 1,225}\). While this still gets the right answer, the inefficient path often leads to arithmetic mistakes.

Alternatively, they might incorrectly combine terms as \(\mathrm{p + 0.4p = 0.4p}\) or \(\mathrm{p + 0.4p = 1.4}\), leading to wrong intermediate calculations.

Second Most Common Error:

Poor SIMPLIFY execution: Students make arithmetic errors in the final calculations. They might correctly get to \(\mathrm{35 \times 35}\) but miscalculate this multiplication, or they might make errors when calculating \(\mathrm{1.4 \times 25}\) or \(\mathrm{60 - 25}\). These calculation errors lead them to select one of the other answer choices like Choice B (1,050) or Choice C (1,155).

The Bottom Line:

This problem tests whether students can systematically work through function evaluation while maintaining accuracy in both algebraic simplification and arithmetic calculations. The key insight is to simplify first, then substitute—not the other way around.