SOLUTION: i'am having problem solving this equation i have asked twice today and have not gotten a reply back. Maybe you do not understand it either. I know that 2w+2l= perimeter, but how do

Question 55406: i'am having problem solving this equation i have asked twice today and have not gotten a reply back. Maybe you do not understand it either. I know that 2w+2l= perimeter, but how do i apply it to this word problem .
A garden plot has an area of 120 square meters. The measure of the length and width are whole numbers. The length and width of the garden were selected so that the least amount of fencing was required to enclose the garden. What are the dimensions of this garden.
Where do i go from 2w+2l=120
Found 2 solutions by stanbon, Cintchr:
Answer by stanbon(48568) (Show Source):
You can put this solution on YOUR website!
You are mixing perimeter and area.
You have area = 120
Let length = l and width = w
lw=120
You are told the numbers are whole so you look at candidates.
1 and 120 Area = 120; 1+120= 121
2 and 60 Area = 120; 2+60 = 62
3 and 40 Area = 120; 3+40= 43
.....
Keep this up till you find the mimimum sum.
Ans:
10 and 12 Area = 120; 10+12 = 22
That gives you the mimimum
Cheers,
Stan H.

Answer by Cintchr(475) (Show Source):
You can put this solution on YOUR website!
and
If the area = 120 then list out the possible dimensions :
AREA ............. Perimiter
1 * 120 .......... 2(1+120) = 242
2 * 60 ........... 2(2+60) = 124
3 * 40 ........... 2(3+40) = 86
4 * 30 ........... 2(4+30) = 68
5 * 24 ........... 2(5+24) = 58
6 * 20 ........... 2(6+20) = 52
8 * 15 ........... 2(8+15) = 46
10 * 12 .......... 2(10+12) = 44
For future reference .... the square, or the rectangle that is as close to a square as possible, will have the smallest perimiter.