Answer:
[tex]length = 51.96\;m, breadth = 51.96\;m[/tex]
Step-by-step explanation:
Rectangular field:
[tex]Area = l \times b[/tex] unit^2
[tex]2700 = l \times b[/tex]
[tex]b=\dfrac{2700}{l} [/tex] - Eq(A)
The cost of the fence depends on the side lengths of the rectangular fields (the perimeter) plus an additional fence down the middle
[tex]Total\;\;Perimeter = 2l + 2b + l[/tex] units
[tex]Cost = 2(6l) + 2(9b) + (6l) [/tex] $
[tex]Cost = 18l + 18b[/tex] - Eq(B)
Since we want the minimum cost, we need to either differentiate the above equation OR just plot it.
but we need to have only variable on both sides of the equal (=) sign.
so we can substitute Eq(A) in Eq(B)
[tex]Cost = 18l + 18\left(\dfrac{2700}{l} \right) [/tex]
now just simplify it!
[tex]Cost = 18l + \dfrac{48600}{l}[/tex]
the figure below states that at L = 51.96 meters the cost is least.
by plotting this equation, we'll be able to clearly see the length at which the cost is least! Anywhere else will increase the cost!
We can use [tex]l = 51.96[/tex] to find [tex]b[/tex] from Eq(A)
[tex]b=\dfrac{2700}{l} [/tex]
[tex]b=\dfrac{2700}{51.96} [/tex]
[tex]b= 51.96 [/tex] meters
The dimensions of the rectangle at which the cost of the fence is minimum is::
[tex]length = 51.96\;m, breadth = 51.96\;m[/tex]
