SOLUTION: You are designing a rectangular enclosure with 3 rectangular sections separated by parallel walls. You have 4,000 feet of fencing. What is the maximum area that can be enclosed?

Algebra ->  Customizable Word Problem Solvers  -> Geometry -> SOLUTION: You are designing a rectangular enclosure with 3 rectangular sections separated by parallel walls. You have 4,000 feet of fencing. What is the maximum area that can be enclosed?       Log On

Ad: Over 600 Algebra Word Problems at edhelper.com


    

Question 1096893: You are designing a rectangular enclosure with 3 rectangular sections separated by parallel walls. You have 4,000 feet of fencing. What is the maximum area that can be enclosed?
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Join the 3 sections together along their +W+ sides
Each section is an +L+ by +W+ rectangle
Adding up the side lengths, let +K+ =
total length of fencing needed
+K+=+4W+%2B+6L+
( note that 2 of the W sides are internal sides )
Let +A+ = total area
+A+=+3L%2AW+
-----------------------
+K+=+4000+
+4W+%2B+6L+=+4000+ ft
+L+=+%28+4000+-+4W+%29+%2F+6+
+L+=+%28+2000+-+2W+%29+%2F+3+
--------------------------
Substituting:
+A+=+%28+2000+-+2W+%29%2AW+
+A+=+-2W%5E2+%2B+2000W+
-------------------------
The W-value of +A%5Bmax%5D+ is:
+W%5Bmax%5D+=+-b%2F%282a%29+
+a+=+-2+
+b+=+2000+
+W%5Bmax%5D+=+-2000%2F%282%2A%28-2%29%29+
+W%5Bmax%5D+=+500+
Plug this result back into equation
+A%5Bmax%5D+=+-2%2A500%5E2+%2B+2000%2A500+
+A%5Bmax%5D+=+-500000+%2B+1000000+
+A%5Bmax%5D+=+500000+
-------------------------
The maximum area is 500,000 ft2
---------------------------------
check:
+3L+=+A%2FW+
+3L+=+500000%2F500+
+3L+=+1000+
and
+K+=+4W+%2B+6L+
+K+=+4%2A500+%2B+6%2A%281000%2F3%29+
+K+=+2000+%2B+2000+
+K+=+4000+ ( given )
+4000+=+4000+
OK
Check the math & get
another opinion if needed