Question 109832: A developer wants to enclose a rectangular lot that borders a city street. If the developer has 248 ft. of fence and does not fence the side bordering the street, what is the largest area that can be enclosed.
Answer by bucky(2189)   (Show Source):
You can put this solution on YOUR website! Let's begin by picturing what the developer will do. Taking his roll of fencing, he will
start at the edge of the street and will go perpendicular to the street for some unknown distance.
Call that distance x. Then he will turn the fence line 90 degrees and go parallel to the
street some unknown distance. Call that distance y. The he will turn the fence line by
90 degrees again and will head back to the street. If you are picturing this correctly
you will see that the distance to return to the edge of the street must again be x, the same
distance as he originally went away from the street.
.
Now we can see that in using up all 248 feet of fencing, he goes a distance x away from the
street, a distance y parallel to the street, and a distance x back to the street.
We can write this is equation form as:
.
x + y + x = 248
.
and combine the two x terms to reduce it to:
.
2x + y = 248
.
Then we can solve for y in terms of x by subtracting 2x from both sides of this equation
to get:
.
y = 248 - 2x
.
So in terms of x we now have that one side of the rectangle is x and the other dimension
of the rectangle is 248 - 2x.
.
The area (A) of this rectangle is just the product of these two dimensions. (Area of a
rectangle is the product of length times width).
.
So for this rectangle we can write the Area equation as:
.
A = x(248 - 2x)
.
At this point we can do some "thought analysis." If you multiply out these two terms you
will get the quadratic equation A = 248x - 2x^2. The graph of a quadratic equation
is a parabola, and in this case the minus sign on the x^2 term tells you that as you move
from left to right on the graph, the parabola rises to a peak and then drops down the
further you go ... similar to the arc traced by a football pass.
.
Since we are graphing the Area (vertical axis) against x (horizontal axis) we can tell
that at some point on the graph it comes from below the x-axis, crosses it and goes positive.
How can we tell? Easy. The area has to be a positive value, so the graph must be above the
x-axis somewhere. The area rises to a peak and then starts to drop off as x keeps increasing.
At some larger value of x the graph again crosses the x-axis on its way down.
.
Where the graph crosses the x-axis the value of the area (A) must be zero. Therefore,
our equation becomes:
.
0 = x(248 - 2x)
.
Notice that this equation will be true if either factor equals zero. So our graph crosses
the x-axis when x = 0 (on its way up) and again when (248 - 2x) = 0 (on its way down).
Solving the second factor for x :
.
248 - 2x = 0
.
Subtract 248 from both sides:
.
-2x = -248
.
Then divide both sides by -2 to get:
.
x = -248/-2 = 124
.
So the graph crosses the x-axis when x = 0 and when x = 124. Because the parabola
is symmetrical the peak will be reached halfway between these two values, and the midway
value of x between x = 0 and x = 124 occurs when x = 124/2 = 62 feet.
.
So the maximum area will occur when x (the distance from the edge of the street to the
first corner in the fence) is 62 feet. Going back to our original "picture" of the fence,
we can get that the length of the fence (248 feet) is now comprised of 62 feet perpendicular
to the street, then y feet parallel to the street, and finally another 62 feet to get
back to the street edge. So y must be 248 minus the two x-sides or 248 - 2(62) which turns
out to be 124 feet. The area enclosed by the fence will be 62 times 124 which is 7688 sq ft.
.
You can check this out by supposing that x was 61 feet. If you work it out, the length of
the fence parallel to the street will be 126 feet, and the area enclosed will therefore
be 61 times 126 or 7686 square feet ... a little smaller than the area if x is 62 feet.
.
Similarly, if x is 63 feet, then the length of the fence parallel to the street will work
out to be 122 feet, and the area enclosed will be 63 times 122 which will again result in 7686
square feet. Again this area is slightly smaller than the area we got when x was 62 feet.
.
Now that we have done this analysis (mainly to get a feel for what is going on here) we
can return to the equation:
.
A = 248x - 2x^2
.
Set the area equal to zero and rearrange this equation into the quadratic form of:
.
-2x^2 + 248x = 0
.
From the quadratic formula we know that the first term for the answer to x is:
.
-b/(2*a)
.
In this standard form we can see a = -2 and b = 248. Plugging these values into the term
.
-b/(2*a)
.
results in:
.
-(248)/(2*-2) = -248/-4 = 62
.
So using our quadratic equation and -b/(2*a) also gives us the answer 62 feet. Just another
way of doing the problem.
.
In summary, the maximum area that can be enclosed is 7688 sq ft.
.
Hope all this discussion isn't too confusing and gives you some sense of the problem and
how to work it.
.
|
|
|
