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:

Answers

Answer 1
Answer:

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

P = 300m --- the perimeter

L \to Length

W \to Width

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

P = 2L + W

So, we have:

2L + W=300

Make W the subject

W = 300 - 2L

The area of the farm is:

A = L * W

Substitute W = 300 - 2L

A = L * (300 - 2L)

A = 300L - 2L^2

Differentiate, and set the result of the differentiation to 0

A' = 300 - 4L

300 - 4L = 0

Solve for L

4L = 300

Divide by 4

L =75

Solve for W

W = 300 - 2L

W = 300 - 2* 75

W =150

Solve for A

A = L * W

A = 75 * 150

A = 11250

Hence,

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

Read more about areas and perimeters t:

brainly.com/question/11957651

Answer 2
Answer:

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

L+W+W=300

L+2W=300......................(1)

Area=Length*Width

Area=L*W

Plugging the value of L from equation (1)

Area=(300-2W)*W

Area=300W-2W^2

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

Means :

(d)/(dW)Area= 300-4W = 0

So, So, W Width = (300)/(4)=75

So, L Length=300-2*75=150

So, Largest area= 75*150=11250   :    Answer


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yw egen2020