Answer:
Therefore the dimension of the rectangular piece of land is 375 m by 250 m.
Step-by-step explanation:
Given that,
Two opposite sides of rectangular land will be fenced using standard fencing that costs $6 per m while the other two side will require heavy duty fencing that costs $9 per m.
Let the length and width of the rectangular x and y respectively.
Then total cost for fencing = ![\$(2x\times 6+2y\times 9)](https://tex.z-dn.net/?f=%5C%24%282x%5Ctimes%206%2B2y%5Ctimes%209%29)
=$(12x+18y)
Then,
12x+18y= 9,000
⇒ 2x+3y=1,500 [ cancel out common factor 6]
⇒3y= 1500 - 2x
....(1)
The area of the rectangular land is = Length × width
= xy
Let,
A=xy
Putting the value of y
![A=x.\frac{1500-2x}{3}](https://tex.z-dn.net/?f=A%3Dx.%5Cfrac%7B1500-2x%7D%7B3%7D)
![\Rightarrow A=\frac13(1500x-2x^2)](https://tex.z-dn.net/?f=%5CRightarrow%20A%3D%5Cfrac13%281500x-2x%5E2%29)
Differentiating with respect to x
![A'=\frac13(1500-4x)](https://tex.z-dn.net/?f=A%27%3D%5Cfrac13%281500-4x%29)
Again differentiating with respect to x
![A''=-\frac43](https://tex.z-dn.net/?f=A%27%27%3D-%5Cfrac43)
Now we have to determine the critical value of A.
For critical value set A'=0
![\frac13(1500-4x)=0](https://tex.z-dn.net/?f=%5Cfrac13%281500-4x%29%3D0)
⇒1500-4x=0
![\Rightarrow x=\frac{1500}{4}](https://tex.z-dn.net/?f=%5CRightarrow%20x%3D%5Cfrac%7B1500%7D%7B4%7D)
m
Since,
. at x=375 m the area of the rectangular piece of land is maximum.
From equation (1) we get when x=375
![y=\frac{1500-2\times 375}{3}](https://tex.z-dn.net/?f=y%3D%5Cfrac%7B1500-2%5Ctimes%20375%7D%7B3%7D)
⇒y=250 m
Therefore the dimension of the rectangular piece of land is 375 m by 250 m.