Let the dimensions of the rectangular garden be:
- Length = \( l \) meters
- Width = \( w \) meters
The area of the garden is given as \( 100 \, m^2 \), so:
\( l \cdot w = 100 \)
(1) \( l = \frac{100}{w} \)
The farmer fences only three sides of the garden because the fourth side is a compound wall. The total length of the barbed wire available is \( 30 \, m \), so:
\( l + 2w = 30 \)
(2) \( l = 30 - 2w \)
### Solving the equations
From equation (1), we have \( l = \frac{100}{w} \).
From equation (2), \( l = 30 - 2w \).
Equating the two expressions for \( l \):
\( \frac{100}{w} = 30 - 2w \)
Multiply through by \( w \) to eliminate the denominator:
\( 100 = 30w - 2w^2 \)
Rearrange into standard quadratic form:
\( 2w^2 - 30w + 100 = 0 \)
\( w^2 - 15w + 50 = 0 \)
Solve this quadratic equation using the quadratic formula:
\( w = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = -15 \), \( c = 50 \).
\( w = \frac{-(-15) \pm \sqrt{(-15)^2 - 4(1)(50)}}{2(1)} \)
\( w = \frac{15 \pm \sqrt{225 - 200}}{2} \)
\( w = \frac{15 \pm \sqrt{25}}{2} \)
\( w = \frac{15 \pm 5}{2} \)
The two possible values for \( w \) are:
\( w = \frac{15 + 5}{2} = 10 \)
\( w = \frac{15 - 5}{2} = 5 \)
### Corresponding lengths
If \( w = 10 \):
\( l = \frac{100}{10} = 10 \)
If \( w = 5 \):
\( l = \frac{100}{5} = 20 \)
### Verification
1. For \( l = 10, w = 10 \): \( l + 2w = 10 + 2(10) = 30 \). Satisfies the fencing condition.
2. For \( l = 20, w = 5 \): \( l + 2w = 20 + 2(5) = 30 \). Satisfies the fencing condition.
### Conclusion
The farmer has two possible sets of dimensions for the garden:
1. Length = 10 m, Width = 10 m (a square garden).
2. Length = 20 m, Width = 5 m (a rectangular garden).
