Farmer Ed has 350 meters of fencing and wants to enclose a rectangle mat plot that borders on a river. If farmer Ed does not fence the side along the river find the length and width of the plot that will maximize the area. What is the largest area that can be enclosed

We have been given that Farmer Ed has 350 meters of fencing and wants to enclose a rectangle mat plot that borders on a river. Farmer Ed does not fence the side along the river.

First of all we will make a relevant graph to represent the perimeter of fencing for 3 sides as shown in the attachment.

Since Ed will not fence the side along the river, so fencing is needed for 3 sides and perimeter of the fencing would be [tex]x+x+y=2x+y[/tex].

Now we will equate perimeter with 350 as:

[tex]2x+y=350[/tex]

[tex]y=350-2x[/tex]

We know that area of the rectangle is width times length.

[tex]A=x\cdot y[/tex]

Upon substituting the value of y in area equation, we will get:

[tex]A=x\cdot (350-2x)[/tex]

[tex]A(x)=350x-2x^2[/tex]

Now we will find derivative of area function as:

[tex]A'(x)=\frac{d}{dx}(350x)-\frac{d}{dx}(2x^2)[/tex]

[tex]A'(x)=350-4x[/tex]

Now we will equate derivative with 0 and solve for x as:

[tex]0=350-4x[/tex]

[tex]4x=350[/tex]

[tex]x=\frac{350}{4}=87.5[/tex]

Upon substituting [tex]x=87.5[/tex] in equation [tex]y=350-2x[/tex], we will get:

[tex]y=350-2x\Rightarrow 350-2(87.5)=350-175=175[/tex]

Therefore, the length of 175 meters and width of 87.5 meters will maximize the area.

Upon substituting [tex]x=87.5[/tex] in area function, we will get:

[tex]A(87.5)=350(87.5)-2(87.5)^2[/tex]

[tex]A(87.5)=30625-2(7656.25)[/tex]

[tex]A(87.5)=30625-15312.5[/tex]

[tex]A(87.5)=15312.5[/tex]

Therefore, the largest possible area enclosed would be 15312.5 square meters.

Vespertilio Vespertilio · Answer 2 · 2020-04-14T15:36:26+02:00

Answer:

length = 175 m, width = 87.5 m

Step-by-step explanation:

Total length of fence = 350 m

Let L is the length of the plot and W is the width of the plot.

Area of the rectangular plot = length x width

A = L x W .... (1)

fence is along the three sides

So, L + W + W = 350

L + 2W = 350

L = 350 - 2W

Put in equation (1)

A = (350 - 2 W) x W

A = 350 W - 2 W²

Differentiate with respect to W:

dA/dW = 350 - 4 W

Put it equal to zero for maxima and minima:

350 - 4 W = 0

W = 87.5 m

So, L = 350 - 2 x 87.5 = 175 m

Thus, the length is 175 m and the width is 87.5 m.