MATH SOLVE

4 months ago

Q:
# Farmer Ed has 950 meters 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?

Accepted Solution

A:

The perimeter for this case is given by:
[tex] P = y + 2x

[/tex] Substituting values we have: [tex] y + 2x = 950

[/tex] The area is given by: [tex] A = x * y

[/tex] Writing the area based on a variable we have: [tex] A (x) = x * (950 - 2x)

A (x) = -2x ^ 2 + 950x

[/tex] We derive the area to obtain the maximum of the function: [tex] A '(x) = -4x + 950

[/tex] We equal zero and clear x: [tex] -4x + 950 = 0

4x = 950

x = 950/4

x = 237.5

[/tex] Then, the other dimension is given by: [tex] y = 950 - 2x

y = 950 - 2 * (237.5)

y = 475

[/tex] Finally the maximum area is: [tex] A = (237.5) * (475)

A = 112812.5 m ^ 2

[/tex] Answer: The length and width of the plot that will maximize the area are: [tex] x = 237.5 m

y = 475 m

[/tex] The largest area that can be enclosed is: [tex] A = 112812.5 m ^ 2 [/tex]

[/tex] Substituting values we have: [tex] y + 2x = 950

[/tex] The area is given by: [tex] A = x * y

[/tex] Writing the area based on a variable we have: [tex] A (x) = x * (950 - 2x)

A (x) = -2x ^ 2 + 950x

[/tex] We derive the area to obtain the maximum of the function: [tex] A '(x) = -4x + 950

[/tex] We equal zero and clear x: [tex] -4x + 950 = 0

4x = 950

x = 950/4

x = 237.5

[/tex] Then, the other dimension is given by: [tex] y = 950 - 2x

y = 950 - 2 * (237.5)

y = 475

[/tex] Finally the maximum area is: [tex] A = (237.5) * (475)

A = 112812.5 m ^ 2

[/tex] Answer: The length and width of the plot that will maximize the area are: [tex] x = 237.5 m

y = 475 m

[/tex] The largest area that can be enclosed is: [tex] A = 112812.5 m ^ 2 [/tex]