If you have 1200 yards
of fencing available,
what is the largest
rectangular enclosed field
you could make if you only
need to fence three sides
due to a cliff?
So let's assume this red
side here is the cliff.
We want to enclose the
largest rectangular field
with 1200 yards of fencing.
Let's start by labeling our diagram.
Let's go ahead and let the
length of this side here
equal x so the opposite
side would also be x
and then this side here we'll call y
and we're going to form two equations.
One equation that's going to represent
the area of the rectangle
which we want to maximize
and then another equation
called the constraint
dealing with the amount
of fence that we have.
So based upon how we labeled
this diagram, our constraint
would be that x plus y plus x
must equal 1200 yards
or we can say two x plus y
must equal 1200.
Now, the equation that
we're trying to maximize
is the area equation for this rectangle.
So with that rectangle
labeled the way it is,
the area is going to
be equal to x times y.
And we want to maximize the area.
So we know how to determine
the maximum of an equation
that's written in terms of one variable
but here we have A written
in terms of two variables.
So what we'll do is solve
the constraint for x or y
and then perform substitution
into our area equation.
Looking at our constraint,
it's going to be easier if
we solve this equation for y.
The constraint would be y must equal 1200
minus two x
and now we can substitute
1200 minus two x for y
into the area equation and
then we can maximize it.
So the area is going to
equal x times the quantity,
1200 minus two x, and we'll
go ahead and distribute.
We'll have 1200 x
minus two x squared.
So now in order to
maximize our area function,
we'll determine the critical numbers
and then from there we can determine
the value of x that
will maximize the area.
So to determine the critical numbers,
we'll find the derivative
and then determine where's it's
equal to zero or undefined.
The derivative of our area
function would be 1200
minus four x
since the derivative
exists for all values of x.
To determine the critical numbers,
we'll set the sequel to
zero and solve for x.
We'll divide by four.
So we have x equals 300.
Now, even though we only
obtain one critical number,
we should verify that this
will produce a maximum value,
not a minimum value,
and we could do this a couple ways.
If we determine the second derivative,
we could determine if
the function is concave
up or down at this value of x.
So A double prime is
equal to negative four
which means the function
is always concave down
on the entire interval
which means at the critical
number of x equals 300,
we do have a relative maximum.
So when x is equal to 300,
the area is maximized.
Now, let's go ahead and determine
the dimensions of this rectangle
as well as the maximum area.
Our constraint was two
x plus y must equal 1200
and we just determine
that x is equal to 300.
So now, we can determine y
by performing substitution.
Two times 300 plus y
must equal 1200.
So this is 600.
So if we subtract 600 on both sides,
we have y equals 600.
And these are both in yards.
So now, I can determine the maximum area
of this enclosed field.
Remember, the area was equal to x times y
so we'll have 300 yards
times 600 yards
so the maximum area is
going to be equal to 180,000
square yards.
Let's go back and label our sketch.
This was 300 yards
and so was this
and y was 600 yards
and our area was 180,000 square yards.
I hope you found this helpful.
