# 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 separate animals. if there is 60m of fencing...

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 separate animals. if there is 60m of fencing available to build around this corral then what are the dimensions that enclose the max area?

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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.**

To maximize the area of corral which is divides in two equal parts, and uses exactly 60 m of fence to cover its perimeter plus the area dividing the corral the best shape is a circle divided into two semicircles.

Let:

r = radius of this circular corral

Then length of its perimeter plus the line dividing it in two equal parts (semicircles) is, equal to the length of fence. This is given by:

Length of fence = 2*Pi*r + 2r = 60 (given)

==> r(2*Pi + 2) = 60

r(2*3.14159 +2) = 60

r*8.28318 = 60

r = 7.2436 m

The total area covered by this corral will be given by:

Area = Pi*r^2

= 3.14159*7.2436^2 = 164.8382 m^2

Thus the coral will be enclosed by a circular fence of radius 7.2436 m, and divides in two equal part by fence running along a diameter of this circle.

We presume the corral has two identical or similar shapes with apartition chord , the more the area of the circle with the same material of 60 m fencing . In this way the maximisation can not go beyond pi *(60/2pi)^2 = 900/pi = 286.49 m^2.