Question

Austin wants to build a rectangular enclosure for his animals. One side of the pen will be against a barn, so he needs no fence on that side. The other three sides will be enclosed with wire fencing. If Austin has 450 feet of fencing, what are the dimensions that maximize the area of the enclosure? Prove your answer algebraically using a quadratic model.

Answer 1

Answer:

The dimensions of rectangular enclosure are $x=112.5ft$ & $y=200ft$  and maximum area enclosed by it is $22500ft^{2}$.

Step-by-step explanation:

Diagram of the given scenario shown below:

From the question,

Total length of three sides of rectangular enclosure for wire fencing is $450 ft$.

So Let,

           Length of rectangular enclosure be $xft$.

           Breadth of rectangular enclosure be $y ft$.

Now,  

        ⇒                 $x+y+x=450ft$

        ⇒                    $2x+y=450ft$      ................. (1)

            Area of Rectangle = $length\times breadth$

           Area of Rectangular enclosure  = $x\times y$

putting the value of y from equation (1) into equation (2) we get,

           Area of rectangular enclosure = $x(450-2x)$         ............(2)

 

The maximum Area of rectangular enclosure  occurs at the Average of the solutions to this  Quadratic Equation is

               ⇒                    $x(450-2x)=0$

                                    $x=0$       ,     $x=\frac{450}{2}$

                                    $x=0$        ,     $x=112.5$

So, taking the average of the solution

             ⇒                   $x= \frac{225+0}{2}$

             ⇒                   $x=112.5 ft$

Now , putting the value of x in equation (1) and finding the value of y.

         ⇒                    $y= 450-2\times112.5$

         ⇒                      $y= 200ft$

therefore,

Maximum  Area of the rectangular enclosure is $Area=112.5\times200$                                      

                                                                                           $=22500ft^{2}$