Answer:
The least amount of fencing needed for the rectangular pen is 72.19 feet.
Explanation:
The area and perimeter equations of the pen are, respectively:
(1)
(2)
Where:
- Perimeter, in feet.
- Area, in square feet.
- Width, in feet.
- Length, in feet.
Let suppose that total area is known and perimeter must be minimum, then we have a system of two equations with two variables, which is solvable:
From (2):
![y = (A)/(x)](https://img.qammunity.org/2022/formulas/mathematics/high-school/inqkk65kbdh8pzfletjgrvmrey612ts7c0.png)
(2) in (1):
![p = 2\cdot \left(x + (A)/(x)\right)](https://img.qammunity.org/2022/formulas/mathematics/high-school/6mqkgvywxun1gk7tp759ta8493bv8hkzv2.png)
And the first and second derivatives of the expression are, respectively:
(3)
(4)
Then, we perform the First and Second Derivative Test to the function:
First Derivative Test
![2\cdot \left(x - (A)/(x^(2)) \right) = 0](https://img.qammunity.org/2022/formulas/mathematics/high-school/oyfp28c27luhi9jtpedg47rpfjijzh3k28.png)
![2\cdot \left((x^(3)-A)/(x^(2)) \right) = 0](https://img.qammunity.org/2022/formulas/mathematics/high-school/az8jhsfgudqhr0cptsqwy1wwmt7ppawyes.png)
![x^(3) - A = 0](https://img.qammunity.org/2022/formulas/mathematics/high-school/uwkcq122hszoi8l7al2xjvzi12562g5dvx.png)
Given that dimensions of the rectangular pen must positive nonzero variables:
![x^(3) = A](https://img.qammunity.org/2022/formulas/mathematics/high-school/ezdw7t9w3y0o0yb7dskacsccjtcfsowcgi.png)
![x = \sqrt[3]{A}](https://img.qammunity.org/2022/formulas/mathematics/high-school/6ji0ua5xq8c96n3busz6e5guqxmegzw7hc.png)
Second Derivative Test
![p'' = 4](https://img.qammunity.org/2022/formulas/mathematics/high-school/q0f3ff7ceux9xvuyibrkgu8gv3uvnubi1w.png)
In a nutshell, the critical value for the width of the pen leads to a minimum perimeter.
If we know that
, then the value of the perimeter of the rectangular pen is:
![x = \sqrt[3]{169\,ft^(2)}](https://img.qammunity.org/2022/formulas/mathematics/high-school/3q0g2a45oulxad99giycbym351cikxcz1b.png)
![x \approx 5.529\,ft](https://img.qammunity.org/2022/formulas/mathematics/high-school/qgepzhoh5b524yrgomozs35qdaewsvg4w2.png)
By (2):
![y = (A)/(x)](https://img.qammunity.org/2022/formulas/mathematics/high-school/inqkk65kbdh8pzfletjgrvmrey612ts7c0.png)
![y = (169\,ft^(2))/(5.529\,ft)](https://img.qammunity.org/2022/formulas/mathematics/high-school/osyw7h5pera5dzy1us7uvjznnr04rc8zxn.png)
![y = 30.566\,ft](https://img.qammunity.org/2022/formulas/mathematics/high-school/jzy7m3k829aasei2qjoa1lk9w9x7r0gtgp.png)
Lastly, by (1):
![p = 2\cdot (5.529\,ft + 30.566\,ft)](https://img.qammunity.org/2022/formulas/mathematics/high-school/odgv5c67lu0lihifz6tgfuhhqvjb2l5ifr.png)
![p = 72.19\,ft](https://img.qammunity.org/2022/formulas/mathematics/high-school/uw9fwq1808q91fgh7ibe7t3hmuafnctxmc.png)
The least amount of fencing needed for the rectangular pen is 72.19 feet.