Answer:
x = 125 ft and y = 250/3 ft
Step-by-step explanation:
Let assume that,
x be the length of the northern part of the fence (parallel to the north wall)
y be the length of the western and eastern pieces of the fence
As well as farmer has $4000 to spend hence we can write,
8x + 8y + 4y = 4000
8x + 12y = 4000
Hence we can say that,
from the above equation we can write
[tex]y=\frac{4000}{12} -\frac{2x}{3}[/tex]
We know that area
A=xy.
[tex]A(x) =\frac{500x}{3}-\frac{2x^2}{3}[/tex]
we can write
[tex]A'(x) = \frac{500}{3} -\frac{4x}{3}[/tex]
equating it 0 we get
[tex]A'(x) = \frac{500}{3} -\frac{4x}{3}=0[/tex]
[tex]x=125[/tex]
Also,
[tex]A"(x) = -\frac{4}{3}[/tex]
which is less than zero.
we can see that A''(x) is always less than 0 hence using second derivative test we can say that x = 125 is a maximum point.
now, solving for y we get [tex]y= \frac{250}{3}[/tex]
Hence we can say that dimensions for the plot that would enclose the most area is,
x = 125 ft and y = 250/3 ft
The dimensions for the plot that would enclose the most area are: [tex]125 ft, \frac{500}{3} ft.[/tex]
Let the length of the rectangular plot be x be the length of the northern part of the fence (parallel to the north wall) and y be the length of the western and eastern pieces of the fence.
As well as farmer has $4000 to spend, Thus
8x + 8y + 4y = 4000
8x + 12y = 4000
[tex]x=\frac{4000-12y}{8}[/tex]
And the area of the rectangular fence is:
[tex]A=xy\\A=\frac{4000-12y}{8}y\\ A(y)=500y-\frac{3}{2} y^{2}\\A'(y)=500-{3} y\\A'(y)=0\\500-{3} y=0\\y=\frac{500}{3}[/tex]
[tex]A"(y)=-3<0[/tex]
we can see that A''(y) is always less than 0 .
Therefore area is maximum.
Now,
[tex]x=\frac{4000-12(\frac{500}{3} )}{8} \\x=250 ft[/tex]
So, the dimensions for the plot that would enclose the maximum area is: [tex]125 ft, \frac{500}{3} ft.[/tex]
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