Question

6.) A kennel owner has 168 feet of fencing with which

to enclose a rectangular region. He wants to subdivide
this region into three smaller rectangles of equal
length. If the total area to be enclosed 832 ft-, what are
the possible dimensions of the region?​

Answer 1

The possible dimensions of the region are

52 ft by 16 ft

32 ft by 26 ft

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

`P=2(x+y)`

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

`P=2(x+y)+2y`

`P=2(x+2y)`

we have

`P=168\ ft`

`168=2(x+2y)`

simplify

`84=(x+2y)` ----->
`x=84-2y` -----> equation A

The total area to be enclosed is 832 ft^2

The area is equal to

`A=xy`

so

`832=xy` ----> equation B

substitute equation A in equation B

`832=(84-2y)y`

`832=84y-2y^2`

`2y^2-84y+832=0`

Solve the quadratic equation by graphing

The solution are y1=16 ft, y2=26 ft

see the attached figure

For

`y=16\ ft\\x=84-2(16)=52\ ft`

For

`y=26\ ft\\x=84-2(26)=32\ ft`

therefore

The possible dimensions of the region are

52 ft by 16 ft

32 ft by 26 ft