SOLUTION: hello can you please help me solve," rectangular garden has a perimeter 66 ft and area 216 ftsquared. how can i find the dimensions of the garden." I know that P=2L +2W and Area= L

Question 661498: hello can you please help me solve," rectangular garden has a perimeter 66 ft and area 216 ftsquared. how can i find the dimensions of the garden." I know that P=2L +2W and Area= LxW can i do L times W = 216 ft squared and is L=2w so is it 2w times w = 216 AND THEN I AM STUCK...please help!
Found 3 solutions by ReadingBoosters, stanbon, josmiceli:
Answer by ReadingBoosters(3246)   (Show Source): You can put this solution on YOUR website!
p = 2l+2w=66 or l = (66-2w)/2
a = lw = 216
using area
[(66 - 2w)/2](w) = 216
Multiply both sides by 2
(66-2w)(w) = 216(2)
66w - 2w^2 = 432
-2w^2 + 66w - 432 = 0
-2(w^2 - 33w + 216)=0
divide both sides by -2
w^2 - 33w + 216 = 0
(w - )(w - ) mixed signs due to + 216, but - 33
Look for multiple of 216 that add to make 33: 24,9
(w-24)(w- 9)
w = 24, w = 9
width or length is 24ft
width or length is 9ft
Answer by stanbon(75887)   (Show Source): You can put this solution on YOUR website!
rectangular garden has a perimeter 66 ft and area 216 ftsquared
Equations:
L + W = 33 ft
L*W = 216 ft^2
------
Substitute for L and solve for "W:
(33-W)*W = 216
-------
-W^2 + 33W - 216 = 0
-----
W^2 - 33W + 216 = 0
---
Factor:
(W-9)(W-24) = 0
------
If W = 9 ft, L = 24 ft
If W = 24 ft, L = 9 ft
==========================
Cheers,
Stan H.
Answer by josmiceli(19441)   (Show Source): You can put this solution on YOUR website!

(1) ft
and
ft2
(2)
------------------
(1)
Substitute (1) into (2)
(2)
(2)
(2)
Solve using the quadratic equation

and

Note that
I'll say , and
(1)
(1)
(1)
The width is 9' and the length is 24'

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