MATH SOLVE

5 months ago

Q:
# A campground owner plans to enclose a rectangular field adjacent to a river. the owner wants the field to contain 180,000 square meters. no fencing is required along the river. what dimensions will use the least amount of fencing?

Accepted Solution

A:

Calling x and y the two sizes of the rectangular field, the problem consists in finding the minimum values of x and y that give an area of [tex]A=180000 m^2[/tex].

The area is the product between the two sizes:

[tex]A=xy[/tex] (1)

While the perimeter is twice the sum of the two sizes:

[tex]p=2(x+y)[/tex] (2)

From (1) we can write

[tex]y= \frac{A}{x} [/tex]

and we can substitute it into (2):

[tex]p=2(x+ \frac{A}{x})=2x+2 \frac{A}{x} [/tex]

To find the minimum value of the perimeter, we have to calculate its derivative and put it equal to zero:

[tex]p'(x)=0[/tex]

The derivative of the perimeter is

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

If we require p'(x)=0, we find

[tex]x^2=A[/tex]

[tex]x= \sqrt{A} = \sqrt{180000 m^2}=424.26 m [/tex]

And the other side is

[tex]y= \frac{A}{x}= \frac{180000 m^2}{424.26 m} =424.26 m[/tex]

This means that the dimensions that require the minimum amoutn of fencing are (424.26 m, 424.26 m), so it corresponds to a square field.

The area is the product between the two sizes:

[tex]A=xy[/tex] (1)

While the perimeter is twice the sum of the two sizes:

[tex]p=2(x+y)[/tex] (2)

From (1) we can write

[tex]y= \frac{A}{x} [/tex]

and we can substitute it into (2):

[tex]p=2(x+ \frac{A}{x})=2x+2 \frac{A}{x} [/tex]

To find the minimum value of the perimeter, we have to calculate its derivative and put it equal to zero:

[tex]p'(x)=0[/tex]

The derivative of the perimeter is

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

If we require p'(x)=0, we find

[tex]x^2=A[/tex]

[tex]x= \sqrt{A} = \sqrt{180000 m^2}=424.26 m [/tex]

And the other side is

[tex]y= \frac{A}{x}= \frac{180000 m^2}{424.26 m} =424.26 m[/tex]

This means that the dimensions that require the minimum amoutn of fencing are (424.26 m, 424.26 m), so it corresponds to a square field.