Farmer Ed has 300meters of​ fencing, and wants to enclose a rectangular 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?

Width
Length
Largest area:

Respuesta :

Answer:

Step-by-step explanation:

Alright, lets get started.

Suppose the width is W.

Suppose the length is L.

Total fence is given as 300, so

[tex]L+W+W=300[/tex]

[tex]L+2W=300[/tex]......................(1)

[tex]Area=Length*Width[/tex]

[tex]Area=L*W[/tex]

Plugging the value of L from equation (1)

[tex]Area=(300-2W)*W[/tex]

[tex]Area=300W-2W^2[/tex]

For Area to be maximized, derivative of area should be equal to zero.

Means :

[tex]\frac{d}{dW}Area= 300-4W = 0[/tex]

So, [tex]So, W Width = \frac{300}{4}=75[/tex]

So, [tex]L Length=300-2*75=150[/tex]

So, Largest area= [tex]75*150=11250[/tex]   :    Answer

Ver imagen itemderby

The area of the farm is the amount of space on the farm.

  • The width of the farm is 150 meters
  • The length is 75 meters
  • The largest area is 11250 square meters

Given

[tex]P = 300m[/tex] --- the perimeter

[tex]L \to Length[/tex]

[tex]W \to Width[/tex]

Because one side of the farm is a river, the perimeter is calculated as:

[tex]P = 2L + W[/tex]

So, we have:

[tex]2L + W=300[/tex]

Make W the subject

[tex]W = 300 - 2L[/tex]

The area of the farm is:

[tex]A = L \times W[/tex]

Substitute [tex]W = 300 - 2L[/tex]

[tex]A = L \times (300 - 2L)[/tex]

[tex]A = 300L - 2L^2[/tex]

Differentiate, and set the result of the differentiation to 0

[tex]A' = 300 - 4L[/tex]

[tex]300 - 4L = 0[/tex]

Solve for L

[tex]4L = 300[/tex]

Divide by 4

[tex]L =75[/tex]

Solve for W

[tex]W = 300 - 2L[/tex]

[tex]W = 300 - 2\times 75[/tex]

[tex]W =150[/tex]

Solve for A

[tex]A = L \times W[/tex]

[tex]A = 75 \times 150[/tex]

[tex]A = 11250[/tex]

Hence,

  • The width of the farm is 150 meters
  • The length is 75 meters
  • The largest area is 11250 square meters

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