Question

Consider the following problem: A farmer with 950 ft of fencing wants to enclose a rectangular area and then divide it into four pens with fencing parallel to one side of the rectangle. What is the largest possible total area of the four pens

Answer 1

Answer:

Largest possible total area of the four pens is $22562.5\:ft^{2}$

Step-by-step explanation:

Assume width as x and length as y. Given that length of fencing is 950 feet which encloses rectangular area which is divided into four pens as shown in diagram ( Refer to attachment),  

So perimeter of the rectangular area as per diagram is given as,

Perimeter = width + width + width + width + width + length+ length

Substituting the value,  

$950=x+x+x+x+x+y+y$ ….1

Now area of rectangular diagram is given as follows,  

$A=xy$  ….2 

Solving equation 1 for y, subtracting 2x from both sides,  

$\dfrac{950-5x}{2}= y$

$475-\dfrac{5x}{2}= y$

Substituting the value in equation 2,  

$A=x\left(475-\dfrac{5x}{2}\right)$

Simplifying

$A=475x-\dfrac{5x^{2}}{2}$

To find the largest possible area, differentiate A with respect to x,  

$\dfrac{dA}{dx}=\dfrac{d}{dx}\left(475x-\dfrac{5x^{2}}{2}\right)$

Applying sum rule of derivative,  

$\dfrac{dA}{dx}=\dfrac{d}{dx}\left(475x\right)-\dfrac{d}{dx}\left(\dfrac{5x^{2}}{2}\right)$

Applying constant multiple rule of derivative,  

$\dfrac{dA}{dx}=475\dfrac{d}{dx}\left(x\right)-\dfrac{5}{2}\dfrac{d}{dx}\left(x^{2}\right)$

Applying power rule of derivative,

$\dfrac{dA}{dx}=475\left(1x^{1-1}\right)-\dfrac{5}{2}\left(2x^{2-1}\right)$

$\dfrac{dA}{dx}=475\left(x\right)-\dfrac{5}{2}\left(2x\right)$

$\dfrac{dA}{dx}=475-5x$

Now find the critical number by solving  as follows,

$\dfrac{dA}{dx}=0$

$475-5x =0$

$475=5x$

$95=x$

Since there is only one critical point, directly substitute the value of x into equation of A,

$A=475\left(95\right)-\dfrac{5\left(95\right)^{2}}{2}$

Simplifying,  

$A=45125-\dfrac{45125}{2}$

$A=\dfrac{45125}{2}$

So, the largest possible area is $22562.5\:ft^{2}$