Answer:
3 mi and [tex]\dfrac{4}{3}[/tex] mi, 1 mi and 4 mi
Step-by-step explanation:
Length of one of rectangles = x
Breadth of one of the rectangles = y
Total enclosed area = [tex]8\ \text{mi}^2[/tex]
Fence available = 16 miles
Total area is given by
[tex]2xy=8\\\Rightarrow y=\dfrac{4}{x}[/tex]
Perimeter of the fence
[tex]2\times 2x+3y=16\\\Rightarrow 4x+3y=16\\\Rightarrow 4x+3\dfrac{4}{x}=16\\\Rightarrow 4x+\dfrac{12}{x}=16\\\Rightarrow 4x^2-16x+12=0\\\Rightarrow x=\frac{-\left(-16\right)\pm \sqrt{\left(-16\right)^2-4\times \:4\times \:12}}{2\times \:4}\\\Rightarrow x=3,1[/tex]
If [tex]x=3[/tex]
[tex]y=\dfrac{4}{x}=\dfrac{4}{3}[/tex]
If [tex]x=1[/tex]
[tex]y=\dfrac{4}{x}=4[/tex]
The possible values of length and breadth are 3 mi and [tex]\dfrac{4}{3}[/tex] mi, 1 mi and 4 mi respectively.
