# Thread: What is the value of the dimensions?

1. ## What is the value of the dimensions?

[B]Problem:
A farm has a perimeter of 55 km and has an area of 150 km sq. Find the dimensions of the farm.

i tried to answer on this that
A= l x w P = 2 (l + w)
150 km sq. = l x w 55 km = 2 (l +w)

what is next to here? do i need to manipulate the formula? i'm stuck here.
Please does anybody can help me on this?
Thank you

2. Originally Posted by rcs
[B]Problem:
A farm has a perimeter of 55 km and has an area of 150 km sq. Find the dimensions of the farm.

i tried to answer on this that
A= l x w P = 2 (l + w)
150 km sq. = l x w 55 km = 2 (l +w)

what is next to here? do i need to manipulate the formula? i'm stuck here.
Please does anybody can help me on this?
Thank you
$LW = 150
$

$L = \dfrac{150}{W}$

$2(L+W) = 55$

$2\left(\dfrac{150}{W} + W\right) = 55$

solve for $W$ , then determine $L$

3. thanks a lot

4. Or, the other way: 2(L+ W)= 55 so L+ W= 55/2 and then W= 55/2- L. LW= L(55/2- L)= (55/2)L- L^2= 150.

2(150/w + w) = 55
2(150+w^2w) = 55
(300+ 2w^2) / w = 55
300+ 2w^2 = 55w
2w^2 - 55w + 300 = 0
(2w - 15 ) ( w - 20) = 0
w = 20 or w = 15/2 or 7.5

LW = 150 if w = 20
L = 150/20
L = 7.5

LW = 150 if w = 7.5
L = 150/7.5
L = 20

which of this two values is right/ correct for the L ?

2(150/w + w) = 55
2(150+w^2/w) = 55
(300+ 2w^2) / w = 55
300+ 2w^2 = 55w
2w^2 - 55w + 300 = 0
(2w - 15 ) ( w - 20) = 0
w = 20 or w = 15/2 or 7.5

LW = 150 if w = 20
L = 150/20
L = 7.5

LW = 150 if w = 7.5
L = 150/7.5
L = 20

which of this two values is right/ correct for the L ?

7. They are both correct. Both say that one side has a length of 7.5 km and the other 20 km. You have only switched which side you call "length" and which you call "width".