SOLUTION: Question 9: A rancher has 600 feet of fencing to put around a rectangular field and then subdivide the field into 3 identical smaller rectangular plots by placing two fences parall

Algebra ->  Customizable Word Problem Solvers  -> Misc -> SOLUTION: Question 9: A rancher has 600 feet of fencing to put around a rectangular field and then subdivide the field into 3 identical smaller rectangular plots by placing two fences parall      Log On

Ad: Over 600 Algebra Word Problems at edhelper.com


    

Question 1150427: Question 9: A rancher has 600 feet of fencing to put around a rectangular field and then subdivide the field into 3 identical smaller rectangular plots by placing two fences parallel to one of the field's shorter sides. Find the dimensions that maximize the enclosed area. Write your answers as fractions reduced to lowest terms.
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Let +w+ = the width of the rectangle
+4w+ = the total of the +w+ lengths of fence
+%28+600+-+4w+%29+%2F+2+ = the length of the rectangle
Let +A+ = the total area of the rectangle
----------------------------------------------------------
+A+=+4w%2A%28+%28+600+-+4w+%29+%2F+2+%29+
+A+=+2w%2A%28+600+-+4w+%29+
+A+=+-8w%5E2+%2B+1200w
This is a parabola with a maximum peak
+w%5Bpk%5D+=+-1200%2F%282%2A%28-8%29%29+
+w%5Bpk%5D+=+75+
Plug this result back into equation
+A+=+-8w%5E2+%2B+1200w
+A%5Bmax%5D+=+-8%2A75%5E2+%2B+1200%2A75 ( formula for vertex of parabola )
+A%5Bmax%5D+=+-45000+%2B+90000+
+A%5Bmax%5D+=+45000+
---------------------------------------
+45000%2Fw%5Bpk%5D+=+45000%2F75+
+45000%2Fw%5Bpk%5D+=+600+
---------------------------------------
The dimensions that maximize area are:
75' x 600'
----------------
check:
+graph%28+400%2C+400%2C+-20%2C+200%2C+-5000%2C+50000%2C+-8x%5E2+%2B+1200x+%29+