SOLUTION: A field must be fenced in. There is 212 yards of fencing material to be used. What should the dimensions of the enclosed area be to ensure that it has the largest possible area?

Algebra ->  Coordinate Systems and Linear Equations  -> Linear Equations and Systems Word Problems -> SOLUTION: A field must be fenced in. There is 212 yards of fencing material to be used. What should the dimensions of the enclosed area be to ensure that it has the largest possible area?      Log On


   

Question 1065428: A field must be fenced in. There is 212 yards of fencing material to be used. What should the dimensions of the enclosed area be to ensure that it has the largest possible area?
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Let +W+ = the width
Let +L+ = the length
Using formula for perimeter:
+2W+%2B+2L+=+212+
+W+%2B+L+=+106+
+L+=+106+-+W+
--------------------------
Let +A+ = the area
+A+=+W%2A%28+106+-+W+%29+
+A+=+-W%5E2+%2B+106W+
This is a parabola with the vertex a maximum,
and not a minimum
---------------------
Formula for vertex:
+W%5Bv%5D+=+-b%2F%282a%29+
+a+=+-1+
+b+=+106+
+W%5Bv%5D+=+-106%2F%28+2%2A%28-1%29%29+
+W%5Bv%5D+=+53+
and
+L%5Bv%5D+=+106+-+W+
+L%5Bv%5D+=+106+-+53+
+L%5Bv%5D+=++53+
-------------------------
The enclosed area is maximum when it is 53 yds x 53 yds
check:
+4%2A53+=+212+ yds of fencing
OK
Here's a plot of +A+ and +W+
+graph%28+400%2C+400%2C+-15%2C+150%2C+-350%2C+3500%2C+-x%5E2+%2B+106x+%29+