Can you please explain this is easy steps?A small corral is to be built so that is also has fence splitting the corral into two smaller areas to...

Since there are 60 m of fencing, then the perimeter of the
rectangular fence + the separation fence = 60.

Let x and y
be the sides of the rectangle, let x also be the measure of the fence that separate both
areas.

==>  2x + 2y + x  =
60

==> 3x + 2y =
60

==> 3x = 60
-2y

==> x= 20 -(2/3) y
.........(1)

Now we need to determine the maximum area we
could have from using the 60 m fence.

We know that the area
A is:

A = x*y

But from (1) ,
x= 20 -(2/3) y

==> A =
(20-2/3y)*y

==> A =20y -
(2/3)y^2

Now we need to caclulate the maximum value for
A.

First, since the factor of y^2 is negative, then the
function has a maximum value.

Let us find the derivative's
zero.

A1' = 20 - (4/3)y =
0

==> 20 =
(4/3)y

==> y = 20 * 3/4 =
15

==> x= 20 -(2/3)y = 20 - (2/3)*15= 20 - 10 =
10

Then A = 15*10 = 150
m^2

Then the maximum area we could have from
the 60 m fence is 150 m^2 and that by fencing a square area
15X10.