Answer:
The possible dimensions of the region are
52 ft by 16 ft
32 ft by 26 ft
Step-by-step explanation:
Let
x -----> the length of the rectangular region
y ----> the width of the rectangular region
we know that
The perimeter of a rectangle is equal to
[tex]P=2(x+y)[/tex]
Remember that
A kennel owner wants to subdivide this region into three smaller rectangles of equal length
so
we also have to take into account the divisions
[tex]P=2(x+y)+2y[/tex]
[tex]P=2(x+2y)[/tex]
we have
[tex]P=168\ ft[/tex]
[tex]168=2(x+2y)[/tex]
simplify
[tex]84=(x+2y)[/tex] -----> [tex]x=84-2y[/tex] -----> equation A
The total area to be enclosed is 832 ft^2
The area is equal to
[tex]A=xy[/tex]
so
[tex]832=xy[/tex] ----> equation B
substitute equation A in equation B
[tex]832=(84-2y)y[/tex]
[tex]832=84y-2y^2[/tex]
[tex]2y^2-84y+832=0[/tex]
Solve the quadratic equation by graphing
The solution are y1=16 ft, y2=26 ft
see the attached figure
For
[tex]y=16\ ft\\x=84-2(16)=52\ ft[/tex]
For
[tex]y=26\ ft\\x=84-2(26)=32\ ft[/tex]
therefore
The possible dimensions of the region are
52 ft by 16 ft
32 ft by 26 ft
