Answer:
Largest possible total area of the four pens is [tex] 22562.5\:ft^{2}[/tex]
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,
[tex] 950=x+x+x+x+x+y+y[/tex] ….1
Now area of rectangular diagram is given as follows,
[tex] A=xy [/tex] ….2
Solving equation 1 for y, subtracting 2x from both sides,
[tex]\dfrac{950-5x}{2}= y [/tex]
[tex]475-\dfrac{5x}{2}= y [/tex]
Substituting the value in equation 2,
[tex] A=x\left(475-\dfrac{5x}{2}\right) [/tex]
Simplifying
[tex] A=475x-\dfrac{5x^{2}}{2} [/tex]
To find the largest possible area, differentiate A with respect to x,
[tex] \dfrac{dA}{dx}=\dfrac{d}{dx}\left(475x-\dfrac{5x^{2}}{2}\right)[/tex]
Applying sum rule of derivative,
[tex] \dfrac{dA}{dx}=\dfrac{d}{dx}\left(475x\right)-\dfrac{d}{dx}\left(\dfrac{5x^{2}}{2}\right)[/tex]
Applying constant multiple rule of derivative,
[tex]\dfrac{dA}{dx}=475\dfrac{d}{dx}\left(x\right)-\dfrac{5}{2}\dfrac{d}{dx}\left(x^{2}\right)[/tex]
Applying power rule of derivative,
[tex]\dfrac{dA}{dx}=475\left(1x^{1-1}\right)-\dfrac{5}{2}\left(2x^{2-1}\right)[/tex]
[tex]\dfrac{dA}{dx}=475\left(x\right)-\dfrac{5}{2}\left(2x\right)[/tex]
[tex]\dfrac{dA}{dx}=475-5x[/tex]
Now find the critical number by solving as follows,
[tex]\dfrac{dA}{dx}=0[/tex]
[tex] 475-5x =0[/tex]
[tex] 475=5x [/tex]
[tex] 95=x [/tex]
Since there is only one critical point, directly substitute the value of x into equation of A,
[tex] A=475\left(95\right)-\dfrac{5\left(95\right)^{2}}{2} [/tex]
Simplifying,
[tex] A=45125-\dfrac{45125}{2} [/tex]
[tex] A=\dfrac{45125}{2}[/tex]
So, the largest possible area is [tex] 22562.5\:ft^{2}[/tex]
