The dimensions for the plot that would enclose the most area are a length and a width of 125 feet.
In this question we shall use the first and second derivative tests to determine the <em>optimal</em> dimensions of a rectangular plot of land. The perimeter (
), in feet, and the area of the rectangular plot (
), in square feet, of land are described below:
(1)
(2)
Where:
- Width, in feet.
- Length, in feet.
In addition, the cost of fencing of the rectangular plot (
), in monetary units, is:
(3)
Where
is the fencing unit cost, in monetary units per foot.
Now we apply (2) and (3) in (1):
![p = 2\cdot \left(\frac{A}{l}+l \right)](https://tex.z-dn.net/?f=p%20%3D%202%5Ccdot%20%5Cleft%28%5Cfrac%7BA%7D%7Bl%7D%2Bl%20%5Cright%29)
![\frac{C}{c} = 2\cdot (\frac{A}{l}+l )](https://tex.z-dn.net/?f=%5Cfrac%7BC%7D%7Bc%7D%20%3D%202%5Ccdot%20%28%5Cfrac%7BA%7D%7Bl%7D%2Bl%20%29)
![\frac{C\cdot l}{c} = 2\cdot (A+l^{2})](https://tex.z-dn.net/?f=%5Cfrac%7BC%5Ccdot%20l%7D%7Bc%7D%20%3D%202%5Ccdot%20%28A%2Bl%5E%7B2%7D%29)
![\frac{C\cdot l}{c}-2\cdot l^{2} = 2\cdot A](https://tex.z-dn.net/?f=%5Cfrac%7BC%5Ccdot%20l%7D%7Bc%7D-2%5Ccdot%20l%5E%7B2%7D%20%3D%202%5Ccdot%20A)
(4)
We notice that fencing costs are directly proportional to the area to be fenced. Let suppose that cost is the <em>maximum allowable </em>and we proceed to perform the first and second derivative tests:
FDT
![l = \frac{C}{4\cdot c}](https://tex.z-dn.net/?f=l%20%3D%20%5Cfrac%7BC%7D%7B4%5Ccdot%20c%7D)
SDT
![A'' = -2](https://tex.z-dn.net/?f=A%27%27%20%3D%20-2)
Which means that length leads to a <em>maximum</em> area.
If we know that
and
, then the dimensions of the rectangular plot of land are, respectively:
![l = \frac{4000}{4\cdot (8)}](https://tex.z-dn.net/?f=l%20%3D%20%5Cfrac%7B4000%7D%7B4%5Ccdot%20%288%29%7D)
![l = 125\,ft](https://tex.z-dn.net/?f=l%20%3D%20125%5C%2Cft)
![A = \frac{(4000)\cdot (125)}{2\cdot (8)} -125^{2}](https://tex.z-dn.net/?f=A%20%3D%20%5Cfrac%7B%284000%29%5Ccdot%20%28125%29%7D%7B2%5Ccdot%20%288%29%7D%20-125%5E%7B2%7D)
![A = 15625\,ft^{2}](https://tex.z-dn.net/?f=A%20%3D%2015625%5C%2Cft%5E%7B2%7D)
![w = \frac{15625\,ft^{2}}{125\,ft}](https://tex.z-dn.net/?f=w%20%3D%20%5Cfrac%7B15625%5C%2Cft%5E%7B2%7D%7D%7B125%5C%2Cft%7D)
![w = 125\,ft](https://tex.z-dn.net/?f=w%20%3D%20125%5C%2Cft)
The dimensions for the plot that would enclose the most area are a length and a width of 125 feet.
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