Area (A) = Length (L) x Width (w)
1000 = (w + 15)(w)
1000 = w² + 15w
0 = w² + 15 - 1000
Use Pythagorean Theorem to solve for w:
w = [tex] \frac{-b +/- \sqrt{b^{2} - 4ac}}{2a} = \frac{-15 +/- \sqrt{15^{2} - 4(1)(-1000)}}{2(1)} = \frac{-15 +/- \sqrt{225 + 4000}}{2} [/tex]
= (-15 +/- 65)/2 = [tex] \frac{-15 + 65}{2}, \frac{-15 - 65}{2} = \frac{50}{2}, \frac{-80}{2} = 25, -40[/tex]
Since the width cannot be negative, w = -40 is not valid.
w = 25
L = w + 15 → L = (25) + 15 → L = 40
The question is asking for the perimeter:
Perimeter (P) = 2·L + 2·w
= 2(40) + 2(25)
= 80 + 50
= 130
Answer: If you unwind the fence, it would be 130 ft long
