SOLUTION: A rectangular playground is to be fenced off and divided in two by another fence parallel to one side of the playground. Six hundred and sixty feet of fencing is used. Find the dim

Algebra ->  Surface-area -> SOLUTION: A rectangular playground is to be fenced off and divided in two by another fence parallel to one side of the playground. Six hundred and sixty feet of fencing is used. Find the dim      Log On


   

Question 1077182: A rectangular playground is to be fenced off and divided in two by another fence parallel to one side of the playground. Six hundred and sixty feet of fencing is used. Find the dimensions of the playground that maximizes the total enclosed area. What is the maximum area?
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Let +x+ = the length of each of the 3 sides
that are parallel
+%28+660+-+3x+%29+%2F+2+ = the length of each of
the 2 sides that are perpendicular the the
3 sides that are parallel
---------------------------------------------
Let +A+ = the area of the playground
+A+=+x%2A%28+660+-+3x+%29%2F2+
+A+=+%28-3%2F2%29x%5E2+%2B+330x+
----------------------------
Where the form of the equation is:
+f%28x%29+=+a%2Ax%5E2+%2B+bx+%2B+c+
The maximum x-value is at:
+x%5Bmax%5D+=+-b%2F%282a%29+
+a+=+-3%2F2+
+b+=+330+
+x%5Bmax%5D+=+-330%2F%282%2A%28-3%2F2%29+%29+
+x%5Bmax%5D+=+-330%2F%28-3%29+
+x%5Bmax%5D+=+110+
and
+%28+660+-+3x+%29%2F2+=+%28+660+-+330+%29%2F2+
+%28+660+-+3x+%29%2F2+=+330%2F2+
+%28+660+-+3x+%29%2F2+=+165+
----------------------------
The dimensions to maximize the area are:
110 x 165
--------------------------------------
The max area is:
+110%2A165+=+18150+
18,150 ft2
------------------------
check the answer:
+A+=+%28-3%2F2%29x%5E2+%2B+330x+
+A%5Bmax%5D+=+%28-3%2F2%29%2A110%5E2+%2B+330%2A110+
+A%5Bmax%5D+=+%28-3%2F2%29%2A12100+%2B+36300+
+A%5Bmax%5D+=+-18150+%2B+36300+
+A%5Bmax%5D+=+18150+
OK