A rectangle has a length that is 15 times its width. The function \(\mathrm{y = (15w)(w)}\) represents this situation, where...

1. TRANSLATE the problem information

• Given information:
  • Rectangle has length that is 15 times its width
  • Function: \(\mathrm{y = (15w)(w)}\)
  • y = area in square feet, \(\mathrm{y \gt 0}\)

• What this tells us: We need to identify what each part of this function represents in terms of the rectangle's dimensions

2. INFER what the variables represent

• Since the length is 15 times the width, if we call the width w, then:
  • Width = \(\mathrm{w}\)
  • Length = \(\mathrm{15 \times w = 15w}\)

• The function \(\mathrm{y = (15w)(w)}\) represents area, and we know \(\mathrm{area\ of\ rectangle = length \times width}\)

3. INFER the meaning of each component

• Looking at \(\mathrm{y = (15w)(w)}\):
  • The first part \(\mathrm{(15w)}\) must be the length
  • The second part \(\mathrm{(w)}\) must be the width
  • Their product \(\mathrm{(y)}\) is the area

• This makes sense because: \(\mathrm{Area = length \times width = (15w) \times w = (15w)(w)}\)

Answer: A. The length of the rectangle, in feet

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak TRANSLATE skill: Students may focus on the fact that y represents area and incorrectly think that since \(\mathrm{15w}\) is part of the area expression, it must also represent area.

They might reason: "The function gives area, and \(\mathrm{15w}\) is a big part of that function, so \(\mathrm{15w}\) must be the area too." This leads them to select Choice B (The area of the rectangle, in square feet).

Second Most Common Error:

Incomplete INFER reasoning: Students might recognize that w is the width but get confused about what happens when you multiply w by 15. They might think "\(\mathrm{15w}\) means 15 more than w" rather than "15 times w."

This conceptual confusion about the difference between addition and multiplication in context could lead them to select Choice C (The difference between the length and the width).

The Bottom Line:

This problem tests whether students can connect the algebraic representation of a function with the real-world quantities it represents. The key insight is recognizing that when \(\mathrm{area = length \times width}\), each factor in the area expression corresponds to a dimension of the rectangle.