SOLUTION: a woman has 40 yards of fencing for her yard. what is the maximum area she can enclose? this has to be extended response.. please help.

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Question 670274: a woman has 40 yards of fencing for her yard. what is the maximum area she can enclose? this has to be extended response.. please help.
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
I already know that the maximum area is
when the yard is a square, but I will prove this.
Let +L+ = the length of a rectangular yard
Let +W+ = the width of this rectangular yard
given:
+2L+%2B+2W+=+40+ yds
( this is the definition of circumference of a rectangle )
Divide both sides by +2+
(1) +L+%2B+W+=+20+
(1) +W+=+20+-+L+
---------------
Let +A+ = the area of the yard
(2) +A+=+L%2AW+ ( also a definition )
substitute (1) into (2)
(2) +A+=+L%2A%28+20+-+L+%29+
(2) +A+=+-L%5E2+%2B+20L+
---------------------
The rule is: when a quadratic equation has the form
+ax%5E2+%2B+b%2Ax+%2B+c+, the max ( or min ) occurs where
the x-co-ordinate is at +-b%2F%282a%29+
In this problem,
+a+=+-1+
+b+=+20+
+-b%2F%282a%29+=+-20%2F%28-2%29+
+-b%2F%282a%29+=+10+
+L%5Bmax%5D+=+10+
and, since
(1) +L+%2B+W+=+20+
(1) +10+%2B+W+=+20+
(1) +W+=+10+
Both the length and width are 10, so this is a
square yard
+A+=+10%2A10+
+A+=+100+ square yards
So, the maximum area is when all the sides are 10 yds
Here's the plot with Area on the vertical axis
and Length on the horizontal
+graph%28+500%2C+500%2C+-5%2C+25%2C+-10%2C+110%2C++-x%5E2+%2B+20x+%29+