SOLUTION: Considering the following problem:A farmer has 2400 ft of fencing and wants to fence off a rectangular field that borders a straight river. He does not need a fence along the river

Algebra ->  Functions -> SOLUTION: Considering the following problem:A farmer has 2400 ft of fencing and wants to fence off a rectangular field that borders a straight river. He does not need a fence along the river      Log On


   



Question 1044598: Considering the following problem:A farmer has 2400 ft of fencing and wants to fence off a rectangular field that borders a straight river. He does not need a fence along the river . What are the dimensions of the largest area that he can fence?
a. Experiment with the problem by drawing several diagrams illustrating the situation. Calculate the area of each configuration, and use your results to estimate the dimensions of the largest possible field.
b. Find a function that models the area of the field in terms of one of its sides.
c. Use your model to solve the problem, and compare with your answer to part (a).

Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Let +s+ = the length of one of the sides which
is perpendicular to the river
+2400+-+2s+ = the length of the side which
is parallel to the river
-------------------------------------------
Let +A+ = the area of the rectangular field in square ft
+A+=+s%2A%28+2400+-+2s+%29+
+A+=+-2s%5E2+%2B+2400s+
--------------------------
If the form of the equation looks like:
+f%28s%29+=+a%2As%5E2+%2B+b%2As+%2B+c+, then the s-value of the
vertex of the parabola is:
+s%5Bv%5D+=+-b%2F%282a%29+
--------------------
In this case:
+a+=+-2+
+b+=+2400+
---------------
+s%5Bv%5D+=+%28+-2400+%29+%2F+%28+2%2A%28-2%29+%29+
+s%5Bv%5D+=+600+
and
+A%5Bv%5D+=+s%2A%28+2400+-+2s+%29+
+A%5Bv%5D+=+600%2A%28+2400+-+1200+%29+
+A%5Bv%5D+=+600%2A1200+
+A%5Bv%5D+=+720000+
The maximum area is 720,000 ft2
------------------------------
Knowing this, you can try a slightly different
shape for the rectangular field, say by making
+s+=+598+
+2400+-+2s+=+2400+-+1196+
+2400+-+2s+=+1204+
+A+=+598%2A1204+
+A+=+719992+
This is clearly a little less than the maximum area
I just got
Now try:
+s+=+602+
+2400+-+2s+=+2400+-+1204+
+2400+-+2s+=+1196+
+A+=+602%2A1196+
+A+=+719992+
Also a little less than maximum area ( same number as before )
------------------
Here's a plot of the equation for area:
+graph%28+600%2C+400%2C+-250%2C+1500%2C+-80000%2C+800000%2C+-2x%5E2+%2B+2400x+%29+