Question

A farmer is building a fence to enclose a rectangular area consisting of two separate regions. The four walls and one additional vertical segment (to separate the regions) are made up of fencing, as shown below. A rectangular area consisting of two separated regions. If the farmer has 162 feet of fencing, what are the dimensions of the region which enclose the maximal areas?

Answer 1

Answer:

The maximal area will be "1093.5 square feet".

Step-by-step explanation:

Let,

Length = L feet

Breadth = b feet

Given Total fencing = 162 feet

According to the question,

 $(2\times L)+(3\times b)=162$

              $2L+3B=162$

                         $L=\frac{162-3b}{2}$

                         $L=81-\frac{3}{2}b$

As we know,

 $Area=Length\times breadth$

          $=(81-\frac{3}{2}b)\times b$

          $=81b-\frac{3}{2}b^2$

Now, we required to decrease or minimize the are. So for extreme points:

 $\frac{dA}{db}=0$

or,

 $\frac{dA}{dB}=\frac{d}{db}(81-\frac{3}{2}b^2 )=0$

       $81-\frac{3}{2}\times 2\times b=0$

                            $b=\frac{81}{3}$

                            $b=27 \ feet$

Now on putting the value of b, we get

 $l=81-\frac{3}{2}\times 27$

   $=81-40.5$

   $=40.5 \ feet$

So that the dimensions will be:

⇒  40.5 feet by 27 feet

Therefore when the dimension are above then the area will be:

=  $81\times 27-\frac{3}{2}\times 27\times 27$

=  $2187-\frac{3}{2}\times 729$

=  $2187-1093.5$

=  $1093.5 \ square \ feet$