Answer:
![\displaystyle A=\frac{1}{192}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20A%3D%5Cfrac%7B1%7D%7B192%7D)
Step-by-step explanation:
<u>Maximization With Derivatives</u>
Given a function of one variable A(x), we can find the maximum or minimum value of A by using the derivatives criterion. If A'(x)=0, then A has a probable maximum or minimum value.
We need to find a function for the area of the pasture. Let's assume the dimensions of the pasture are x and y, and one divider goes parallel to the sides named y, and two dividers go parallel to x.
The two divisions parallel to x have lengths y, thus the fencing will take 4x. The three dividers parallel to y have lengths x, thus the fencing will take 3y.
The amount of fence needed to enclose the external and the internal divisions is
![P=4x+3y](https://tex.z-dn.net/?f=P%3D4x%2B3y)
We know the total fencing is 1/2 miles long, thus
![\displaystyle 4x+3y=\frac{1}{2}](https://tex.z-dn.net/?f=%5Cdisplaystyle%204x%2B3y%3D%5Cfrac%7B1%7D%7B2%7D)
Solving for x
![\displaystyle x=\frac{\frac{1}{2}-3y}{4}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20x%3D%5Cfrac%7B%5Cfrac%7B1%7D%7B2%7D-3y%7D%7B4%7D)
The total area of the pasture is
![A=x.y](https://tex.z-dn.net/?f=A%3Dx.y)
Substituting x
![\displaystyle A=\frac{\frac{1}{2}-3y}{4}.y](https://tex.z-dn.net/?f=%5Cdisplaystyle%20A%3D%5Cfrac%7B%5Cfrac%7B1%7D%7B2%7D-3y%7D%7B4%7D.y)
![\displaystyle A=\frac{\frac{1}{2}y-3y^2}{4}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20A%3D%5Cfrac%7B%5Cfrac%7B1%7D%7B2%7Dy-3y%5E2%7D%7B4%7D)
Differentiating with respect to y
![\displaystyle A'=\frac{\frac{1}{2}-6y}{4}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20A%27%3D%5Cfrac%7B%5Cfrac%7B1%7D%7B2%7D-6y%7D%7B4%7D)
Equate to 0
![\displaystyle \frac{\frac{1}{2}-6y}{4}=0](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cfrac%7B%5Cfrac%7B1%7D%7B2%7D-6y%7D%7B4%7D%3D0)
Solving for y
![\displaystyle y=\frac{1}{12}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20y%3D%5Cfrac%7B1%7D%7B12%7D)
And also
![\displaystyle x=\frac{\frac{1}{2}-3\cdot \frac{1}{12}}{4}=\frac{1}{16}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20x%3D%5Cfrac%7B%5Cfrac%7B1%7D%7B2%7D-3%5Ccdot%20%5Cfrac%7B1%7D%7B12%7D%7D%7B4%7D%3D%5Cfrac%7B1%7D%7B16%7D)
Compute the second derivative
![\displaystyle A''=-\frac{3}{2}.](https://tex.z-dn.net/?f=%5Cdisplaystyle%20A%27%27%3D-%5Cfrac%7B3%7D%7B2%7D.)
Since it's always negative, the point is a maximum
Thus, the maximum area is
![\displaystyle A=\frac{1}{12}\cdot \frac{1}{16}=\frac{1}{192}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20A%3D%5Cfrac%7B1%7D%7B12%7D%5Ccdot%20%5Cfrac%7B1%7D%7B16%7D%3D%5Cfrac%7B1%7D%7B192%7D)