A rectangle has length l and width w (where l and w represent the length and width in some unit),...

1. TRANSLATE the problem information

• Given information:
  • Area: \(\mathrm{lw = 12}\)
  • Perimeter: \(\mathrm{2l + 2w = 14}\)
  • Constraint: \(\mathrm{l \gt w}\)
• What we need to find: the value of w

2. INFER the solution strategy

• We have two equations with two unknowns - this suggests using substitution
• From the perimeter equation, we can solve for one variable in terms of the other
• The simpler approach: solve the perimeter equation for l, then substitute into the area equation

3. SIMPLIFY the perimeter equation

• Start with: \(\mathrm{2l + 2w = 14}\)
• Divide both sides by 2: \(\mathrm{l + w = 7}\)
• Solve for l: \(\mathrm{l = 7 - w}\)

4. SIMPLIFY by substitution into the area equation

• Substitute \(\mathrm{l = 7 - w}\) into \(\mathrm{lw = 12}\):
• \(\mathrm{(7 - w)w = 12}\)
• Expand: \(\mathrm{7w - w^2 = 12}\)
• Rearrange to standard form: \(\mathrm{w^2 - 7w + 12 = 0}\)

5. SIMPLIFY by factoring the quadratic

• Factor \(\mathrm{w^2 - 7w + 12 = 0}\)
• Look for two numbers that multiply to 12 and add to -7: -3 and -4
• \(\mathrm{(w - 3)(w - 4) = 0}\)
• Therefore: \(\mathrm{w = 3}\) or \(\mathrm{w = 4}\)

6. APPLY CONSTRAINTS to select the final answer

• Test \(\mathrm{w = 3}\): If \(\mathrm{w = 3}\), then \(\mathrm{l = 7 - 3 = 4}\), and we check: \(\mathrm{4 \gt 3}\) ✓
• Test \(\mathrm{w = 4}\): If \(\mathrm{w = 4}\), then \(\mathrm{l = 7 - 4 = 3}\), and we check: \(\mathrm{3 \gt 4}\) ✗
• Since we need \(\mathrm{l \gt w}\), we must have \(\mathrm{w = 3}\)

Answer: B. 3

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak APPLY CONSTRAINTS reasoning: Students solve the quadratic correctly to get \(\mathrm{w = 3}\) or \(\mathrm{w = 4}\), but fail to check which solution satisfies \(\mathrm{l \gt w}\). They might randomly choose \(\mathrm{w = 4}\) because it's larger, not realizing this makes \(\mathrm{l = 3}\), which violates \(\mathrm{l \gt w}\).

This may lead them to select Choice C (4).

Second Most Common Error:

Poor INFER strategy selection: Students try to solve both equations simultaneously without substitution, leading to algebraic confusion. They might attempt to use elimination method inappropriately or get lost trying to manipulate both equations at once without a clear strategy.

This leads to confusion and guessing among the answer choices.

The Bottom Line:

This problem tests whether students can systematically work through a constraint optimization problem. The key insight is recognizing that finding both solutions to the quadratic is just the beginning - the constraint \(\mathrm{l \gt w}\) is what determines the final answer.