Answer:
Dimensions of each pen are [tex]x=5\sqrt{5}[/tex] and [tex]y=4\sqrt{5}[/tex].
Step-by-step explanation:
Please find the attachment.
We have been given that a rancher decides to make 4 identical and adjacent rectangular pens against her barn each with an area of [tex]100m^2[/tex].
The area of rectangle is width times length, so we can set an equation as:
[tex]x\cdot y=100...(1)[/tex]
The fence of 4 identical and adjacent rectangular pens will be equal to perimeter of 4 adjacent rectangles as:
[tex]\text{Perimeter}=4x+5y[/tex]
From equation (1), we will get:
[tex]y=\frac{100}{x}[/tex]
Upon substituting this value in perimeter equation, we will get:
[tex]P=4x+5(\frac{100}{x})[/tex]
[tex]P=4x+\frac{500}{x}[/tex]
Now, we will find the first derivative of perimeter equation as:
[tex]P=4x+500x^{-1}[/tex]
[tex]P'=4-500x^{-2}[/tex]
Now, we will equate 1st derivative equal to 0 to find the critical points:
[tex]4-500x^{-2}=0[/tex]
[tex]-\frac{500}{x^{2}}=-8[/tex]
[tex]\frac{500}{x^{2}}=4[/tex]
[tex]4x^{2}=500[/tex]
[tex]x^{2}=125[/tex]
[tex]x=\sqrt{125}[/tex]
[tex]x=5\sqrt{5}[/tex]
Now, we will find 2nd derivative of above equation as:
[tex]P''=0-(-2*500)x^{-2-1}[/tex]
[tex]P''=1000x^{-3}[/tex]
[tex]P''=\frac{1000}{x^3}[/tex]
Now, we will check point [tex]x=5\sqrt{5}[/tex] in 2nd derivative, if it is positive, then x will be a minimum point.
[tex]P''(5\sqrt{5})=\frac{1000}{(5\sqrt{5})^3}[/tex]
[tex]P''(5\sqrt{5})=\frac{1000}{625\sqrt{5}}[/tex]
[tex]P''(5\sqrt{5})=0.71554[/tex]
Since 2nd derivative is positive, so fence will be minimum at [tex]x=5\sqrt{5}[/tex].
Now, we will substitute [tex]x=5\sqrt{5}[/tex] in equation [tex]y=\frac{100}{x}[/tex] to solve for y as:
[tex]y=\frac{100}{5\sqrt{5}}[/tex]
[tex]y=\frac{20}{\sqrt{5}}[/tex]
[tex]y=4\sqrt{5}[/tex]
Therefore, the fence will be minimum at [tex]y=4\sqrt{5}[/tex].
