Answer:
The least amount of fencing needed for the rectangular pen is 72.19 feet.
Step-by-step explanation:
The area and perimeter equations of the pen are, respectively:
[tex]p = 2\cdot (x + y)[/tex] (1)
[tex]A = x\cdot y[/tex] (2)
Where:
[tex]p[/tex] - Perimeter, in feet.
[tex]A[/tex] - Area, in square feet.
[tex]x[/tex] - Width, in feet.
[tex]y[/tex] - 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):
[tex]y = \frac{A}{x}[/tex]
(2) in (1):
[tex]p = 2\cdot \left(x + \frac{A}{x}\right)[/tex]
And the first and second derivatives of the expression are, respectively:
[tex]p' = 2\cdot \left(1 -\frac{A}{x^{2}} \right)[/tex] (3)
[tex]p'' = \frac{4\cdot A}{x^{3}}[/tex] (4)
Then, we perform the First and Second Derivative Test to the function:
First Derivative Test
[tex]2\cdot \left(x - \frac{A}{x^{2}} \right) = 0[/tex]
[tex]2\cdot \left(\frac{x^{3}-A}{x^{2}} \right) = 0[/tex]
[tex]x^{3} - A = 0[/tex]
Given that dimensions of the rectangular pen must positive nonzero variables:
[tex]x^{3} = A[/tex]
[tex]x = \sqrt[3]{A}[/tex]
Second Derivative Test
[tex]p'' = 4[/tex]
In a nutshell, the critical value for the width of the pen leads to a minimum perimeter.
If we know that [tex]A = 169\,ft^{2}[/tex], then the value of the perimeter of the rectangular pen is:
[tex]x = \sqrt[3]{169\,ft^{2}}[/tex]
[tex]x \approx 5.529\,ft[/tex]
By (2):
[tex]y = \frac{A}{x}[/tex]
[tex]y = \frac{169\,ft^{2}}{5.529\,ft}[/tex]
[tex]y = 30.566\,ft[/tex]
Lastly, by (1):
[tex]p = 2\cdot (5.529\,ft + 30.566\,ft)[/tex]
[tex]p = 72.19\,ft[/tex]
The least amount of fencing needed for the rectangular pen is 72.19 feet.
