# A farmer has 300 feet of fencing and wants to enclose a rectangular corral that borders his barn on one side and then divide it into two plots with a fence parallel to one of the sides. ?

## Assuming that the farmer will not fence the side along the barn, what are the lengths of the parts of the fence if the total area enclosed is 4800 square feet Can you please show work thank you

Apr 14, 2017

Each part $= 49 f t$

#### Explanation:

Two plots of $2400 f {t}^{2}$each

$\therefore 2400 f {t}^{2} = \sqrt{2400}$

$= \pm 49 f t$

$5$sides to fence off.

$\therefore 5 \times 49 = 245 f t$
$300 - 245 = 55 f t$ for parallel fence

Apr 14, 2017

Three lengths of $20$ feet and one of $240$ feet
Or
Three fences of $80$ feet and one of $60$ feet

Both options give two corrals with a total area of $4800 f {t}^{2}$

#### Explanation:

The farmer will have 3 fences of the same length perpendicular to the barn and one length parallel to the barn. It does not matter where he places the middle fence, the area of the total rectangle will be the same.

Let the length of each of the 3 fences perpendicular to the barn
be $x$ feet.

The length of the parallel fence will be $\left(300 - 3 x\right)$ feet

The total area enclosed by the fences will be $x \times \left(300 - 3 x\right)$

$A = x \left(300 - 3 x\right) = 4800$

$300 x - 3 {x}^{2} = 4800 \text{ } \leftarrow$ make = 0

$0 = 3 {x}^{2} - 300 x + 4800 \text{ } \leftarrow \div 3$

$0 = {x}^{2} - 100 x + 1600 \text{ } \leftarrow$ find factors

$0 = \left(x - 20\right) \left(x - 80\right)$

Putting each factor equal to $0$ gives $x = 80 , \text{ or } x = 20$

Both these options will work for the farmer:

• If the shorter sides (perpendicular to the barn are $20$ feet:

$20 + 20 + 20 + 240 = 300$feet

Total rectangle has sides of $20$ feet by $240$ feet

$A = 20 \times 240 = 4800 f {t}^{2}$
The middle fence can be placed anywhere along the longer side.

• If the perpendicular sides are each $80$ feet

The side parallel to the barn will be $60$feet
$80 + 80 + 80 + 60 = 240$ feet

$A = 80 \times 60 = 4800 f {t}^{2}$

Again, the middle fence can be placed anywhere along the side parallel to the barn without affecting the amount of fencing used or the fenced area thus obtained.