- Here's a nice application problem
that you see often in
both algebra and calculus.
We want to enclose a
field with some fencing
and we have 100 feet of
fencing to do it with.
The field is supposed
to be 1250 square feet.
One side is already blocked by a mountain.
Now the question is what do
the other three sides need
to be in order to enclose this field.
Let's draw a picture of what's going on
and label some variables.
So here is our mountain.
And it's going to be a rectangular field.
Now, what we need, should
of said it in the problem,
this is going to be a rectangular field,
these two sides will be the
same but this side doesn't
necessarily have to be the same size.
So we can use the same
variable for these two sides,
but we need to use a
different one for this side.
Now we have 100 feet of fencing
that is going to along these
three sides, so X+X+Y or 2X+Y,
gives us our total
fencing which is 100 feet.
Now, what's going to happen
is we also know we need to
enclose 1250 square
feet, so that's the area,
which is X times Y that has to be 1250.
So you have two equations,
each with X and Y.
So in this we're going to
use substitution in order
to get our equation in
terms of one variable.
Let's take this first
one and solve it for Y.
So Y equals 100-2X and we're going
to plug this into this Y down here.
So we have X times 100
minus 2X equals 1250.
We're going to go ahead
and distribute this X
as part of our solving, so
100X - 2X squared equals 1250.
Right now we can put each of these sides
into our graphing calculator
and see where the intercept.
Let's see if we can solve this by hand
by making it a quadratic equal to zero.
I'm going to bring everything
over to the right-hand side.
I'm going to add 2X squared.
That'll give me a positive 2X squared.
I'm going to subtract 100X.
And then I've got the
positive 1250 at the end.
This whole thing is equal to zero.
I need to see where the
zeroes are for this.
I'm going to come up here and
notice all of these are even,
so I can factor a two out.
Might make this thing easier to deal with.
Two times X squared minus
50X plus 625 equals zero.
Now I don't need to worry about the two,
as far as finding the zeroes.
All I gotta do is look at
the part in the parenthesis.
Now what I need are two numbers
that when multiplied together give me 625,
and when added together give me 50.
So this one works out pretty nice because
25 and 25 are such numbers.
Cause 25 times 25 is 625
and 25 plus 25 gives us 50.
We get that the dimension of
sides X has to be 25 feet.
To find y, well remember we already
solved y in terms of x here
so y equals 100 - two times
25 feet or 50, y equals our 50 feet.
Now this one worked out really nice
because it happened to factor.
If it didn't, we would
have gotten to this step
where we had our quadratic equal to zero,
and we could have used
our quadratic formula
or our graphing calculator
to find our solutions.
