SOLUTION: Alright, here is my problem. How do you develop a formula for perimeter for a raectangle that has a width of {{{y mi}}} and {{{Area = 2xy mi^2}}} ?

Algebra ->  Formulas -> SOLUTION: Alright, here is my problem. How do you develop a formula for perimeter for a raectangle that has a width of {{{y mi}}} and {{{Area = 2xy mi^2}}} ?      Log On


   

Question 72861: Alright, here is my problem. How do you develop a formula for perimeter for a raectangle that has a width of y+mi and Area+=+2xy+mi%5E2 ?
Found 2 solutions by jmg, ankor@dixie-net.com:
Answer by jmg(22) About Me  (Show Source):
You can put this solution on YOUR website!
The area formula for a rectangle is Area = length x width
So since the Area = 2xy, and the width is y. That means the length is 2x.
Perimeter of a rectangle = 2(l + w) or 2l + 2w.
So if width = y and the length = 2x. Then fill in the formula.
Perimeter = 2(2x + y) or 2(2x) + 2y which equals 4x + 2y

Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
How do you develop a formula for perimeter for a rectangle that has a width of y mi and Area = 2xy mi^2 ?
:
I understand it as:
Width = y miles, area = 2xy sq miles
:
Length is the area divided by the width:
2xy%2Fy = 2x is the length:
:
Note that y canceled
:
Perimeter = 2L + 2W
:
P = 2(2x) + 2(y)
:
P = 4x + 2y