Question

A rancher decides to make 4 identical and adjacent rectangular pens against her barn each with an area of 100m^2. What are the dimensions of each pen that minimize the amount of fence that must be used?

Answer 1

Answer:

Dimensions of each pen are $x=5\sqrt{5}$ and $y=4\sqrt{5}$.

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 $100m^2$.  

The area of rectangle is width times length, so we can set an equation as:

$x\cdot y=100...(1)$    

The fence of 4 identical and adjacent rectangular pens will be equal to perimeter of 4 adjacent rectangles as:

$\text{Perimeter}=4x+5y$

From equation (1), we will get:

$y=\frac{100}{x}$  

Upon substituting this value in perimeter equation, we will get:

$P=4x+5(\frac{100}{x})$

$P=4x+\frac{500}{x}$  

Now, we will find the first derivative of perimeter equation as:

$P=4x+500x^{-1}$  

$P'=4-500x^{-2}$  

Now, we will equate 1st derivative equal to 0 to find the critical points:

$4-500x^{-2}=0$  

$-\frac{500}{x^{2}}=-8$  

$\frac{500}{x^{2}}=4$  

$4x^{2}=500$  

$x^{2}=125$  

$x=\sqrt{125}$

$x=5\sqrt{5}$  

Now, we will find 2nd derivative of above equation as:

$P''=0-(-2*500)x^{-2-1}$  

$P''=1000x^{-3}$

$P''=\frac{1000}{x^3}$  

Now, we will check point $x=5\sqrt{5}$ in 2nd derivative, if it is positive, then x will be a minimum point.

$P''(5\sqrt{5})=\frac{1000}{(5\sqrt{5})^3}$  

$P''(5\sqrt{5})=\frac{1000}{625\sqrt{5}}$

$P''(5\sqrt{5})=0.71554$

Since 2nd derivative is positive, so fence will be minimum at $x=5\sqrt{5}$.

Now, we will substitute $x=5\sqrt{5}$ in equation $y=\frac{100}{x}$ to solve for y as:

$y=\frac{100}{5\sqrt{5}}$  

$y=\frac{20}{\sqrt{5}}$  

$y=4\sqrt{5}$  

Therefore, the fence will be minimum at $y=4\sqrt{5}$.