SOLUTION: A landscape architect is planning a new nature area in middle of campus. She wants the length to be twice the width, and wants to put a 3foot high retaining wall around the perimet

Algebra ->  Surface-area -> SOLUTION: A landscape architect is planning a new nature area in middle of campus. She wants the length to be twice the width, and wants to put a 3foot high retaining wall around the perimet      Log On


   

Question 1000584: A landscape architect is planning a new nature area in middle of campus. She wants the length to be twice the width, and wants to put a 3foot high retaining wall around the perimeter. There will be 300 total feet wall installed. How wide will this area be?
I think it's α=l*w
But now I'm lost, could someone help me?
Found 2 solutions by josgarithmetic, MathTherapy:
Answer by josgarithmetic(39625) About Me  (Show Source):
You can put this solution on YOUR website!
Perimeter may match the 300 total feet retaining wall.
w width, L length;
L=2w

2L%2B2w=300
and since the question is to find w, just substitute for L.
2%2A2w%2B2w=300
2w%2Bw=150
3w=150
highlight%28w=50%29

Answer by MathTherapy(10555) About Me  (Show Source):
You can put this solution on YOUR website!

A landscape architect is planning a new nature area in middle of campus. She wants the length to be twice the width, and wants to put a 3foot high retaining wall around the perimeter. There will be 300 total feet wall installed. How wide will this area be?
I think it's α=l*w
But now I'm lost, could someone help me? 
Width of nature area: highlight_green%2816%262%2F3%29 ft