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.