Building a fence, what dimensions will require the least amount of fencing?

Question

A farmer plans to enclose a rectangular pasture adjacent to a river. The pasture must contain 405,000 square meters in order to provide enough grass for the herd. No fencing is needed along the river. What dimensions will require the least amount of fencing?

The image shows a rectangle with three sides, the fourth open, because of the river in the problem.

I'm not quite sure where to start with this problem.

Answer 1

So the picture could look like this:

We are trying to optimize the fencing which is equivalent to the perimeter of the 3-sided "rectangle"

p(x) = 2x + y

But we know that the area of the pasture must be 405,000 m² so we can say x•y = 405,000 or y = 405,000/x so, we can substitute "405,000/x" for "y" in the perimeter function.

p(x) = 2x + 405,000/x

To optimize this, we can take the derivative and set it equal to zero:

p(x) = 2x + 405,000x^-1

p'(x) = 2 - 405,000x^-2

2 - 405,000x^-2 = 0

2 = 405,000x^-2

2 = 405,000/x²

2x² = 405,000

x² = 202500

x = √202500 = 450

y = 405,000/450 = 900

So a 450 m x 900 m pasture requires the least amount of fencing.

A 450 m x 900 m pasture would require 450 + 900 + 450 = 1800 meters. If you let the "x" side be 500 m, then the "y" side would need to be 405,000/500 = 810 m making the amount of fencing be 1810 m. If the "x" side was 600, then the "y" side would need to be 405,000/600 = 675 making the total amount of fencing be 1875 m, etc, etc.

Answer 2

Hi Matt! Thanks for asking a great question!

Let's get started! Draw a rectangle (except that it does not have one side of a length drawn). Let x be the width of the rectangle (there should be 2 x's then), and y be the length of the rectangle (there should only be 1 y's, because the other side has a river next to it).

In terms of area, x * y = Area = 405000

In terms of perimeter of the fence, 2x + y = Perimeter. But because we want to simplify the number of variables in the perimeter equation, rearrange the area equation (here, let's make it in terms of x) so that: y = 405000 / x. Substituting that to the perimeter equation would yield:

2x + ( 405000 / x ) = P(x)

We know that we want to minimize the P(x). The rules of calculus states that P(x) can be a relative minimum IF there is a critical number x = c such that P ' (c) = 0 or does not exist, AND P ' (x) changes from negative to positive at c.

Take the derivative of P(x) and get: P ' (x) = 2 - 405000 * x ^ (-2)

The only x value at which P '(x) would not exist is when x = 0, but that does not help because then the area would be 0 * (some y value) = 0, which is not equal to 405000, so it does not meet the criteria and thus can be discarded.

So let's try to find a x value such that P ' (x) = 0.

0 = 2 - 405000 * x ^ (-2) ⇒ 405000 * x ^ (-2) = 2 ⇒ x = 450

Note that when 0 < x < 450, P ' (x) < 0, and when x > 450, P ' (x) > 0. This x value of x = 450 meets all criteria of being a relative minimum. Since there is no other absolute minimum value that is lower than P(450), we may conclude that x = 450 meters and y = 900 meters (simply divide 405000 by 450, look at the area formula) would yield the minimum fencing while still having an area of 405000 squared meters.

>>>>>>>>>>>>>>>>>>>>>>>>>>>(ANSWER) Let me know if you would like additional help! Thanks!