A farmer has 100 meters of fencing to enclose a rectangular area against an existing straight wall, using the fencing...

1. TRANSLATE the problem information

• Given information:
  • Total fencing available: 100 meters
  • Rectangular area against existing wall
  • Fencing needed for only three sides (wall provides the fourth)
  • \(\mathrm{x}\) = length of side parallel to wall

• What this tells us: We need to set up a constraint equation for the fencing usage

2. INFER the constraint relationship

• Since fencing is used for three sides:
  • One side parallel to wall: length \(\mathrm{x}\)
  • Two sides perpendicular to wall: each length \(\mathrm{y}\)
• Constraint equation: \(\mathrm{x + 2y = 100}\)
• This gives us: \(\mathrm{y = \frac{100 - x}{2}}\)

3. TRANSLATE the area into a function

• Area = length × width = \(\mathrm{x \times y}\)
• Substituting: \(\mathrm{A = x \times \frac{100 - x}{2}}\)

4. SIMPLIFY the area expression

• \(\mathrm{A = \frac{x(100 - x)}{2} = \frac{100x - x^2}{2} = 50x - \frac{1}{2}x^2}\)
• This is a quadratic function: \(\mathrm{A = -\frac{1}{2}x^2 + 50x}\)

5. INFER the optimization approach

• Since coefficient of \(\mathrm{x^2}\) is negative \(\mathrm{(-\frac{1}{2})}\), this parabola opens downward
• Maximum area occurs at the vertex

6. SIMPLIFY using the vertex formula

• For quadratic \(\mathrm{ax^2 + bx + c}\), vertex occurs at \(\mathrm{x = -\frac{b}{2a}}\)
• Here: \(\mathrm{a = -\frac{1}{2}, b = 50}\)
• \(\mathrm{x = -\frac{50}{2(-\frac{1}{2})}}\)
• \(\mathrm{= -\frac{50}{-1}}\)
• \(\mathrm{= 50}\)

Answer: C) 50

Why Students Usually Falter on This Problem

Most Common Error Path:

Poor TRANSLATE reasoning: Students misunderstand which sides need fencing and set up the constraint incorrectly. They might think all four sides need fencing, leading to the constraint \(\mathrm{2x + 2y = 100}\), or \(\mathrm{x + y = 50}\). This would give \(\mathrm{y = 50 - x}\) and area \(\mathrm{A = x(50 - x) = 50x - x^2}\), with maximum at \(\mathrm{x = 25}\).

This may lead them to select Choice A (25).

Second Most Common Error:

Weak INFER skill: Students recognize the setup correctly but don't realize this is an optimization problem requiring calculus concepts. Instead, they might test the given answer choices by plugging them in, but without systematic understanding of why one value should be better than others.

This leads to confusion and guessing among the choices.

The Bottom Line:

This problem combines constraint setup with quadratic optimization. Students need to clearly visualize the fencing arrangement AND recognize that finding maximum area requires vertex analysis of the resulting quadratic function.