What is the shape of the rectangle requiring the least amount of fencing?

A rectangular field of given area is to be fenced off along the back of a river. if no fence is needed along the river, what is the shape of the rectangle requiring the least amount of fencing?





-using differential calculus|||Hi,





If the total fence length for the 3 sides is P, and the distance away from the river is x, then the fencing remaining for the length is P - 2x. the area would be x(P - 2x) or Px - 2x虏.





The 1st derivative of this would be P - 4x. Setting this equal to zero to find the maximum gives P - 4x = 0


P = 4x


x = 录P


The distance away from the river to fence will be 录 of the entire perimeter on each side. The other 陆P will be the length that will provide the maximum enclosed area.


So if there was 200 feet of fencing, 录(200) or 50 feet of fencing is the width and 100 feet of fencing parallel to the river would be the length.





I hope that helps!! :-)|||there is not enough information... by not having an area to fence as in a specific number you can only get a generic formula.





or you need the amount of linear feet of fence to find the smallest area to be fenced|||The easy answer is %26quot;rectangle%26quot;, as you ask for the shape. I think you want the actual dimensions.





The length of fencing is L = 2a + b


The area A = a x b





substituting from second eq. back in the first,





L = 2a + A/a





differentiating





dL/da = 2 - A/a2 (a2 is a squared)





setting the slope = 0 to get min and max





0 = 2 - A/a2





A/a2 = 2





2a2 = A





a = +-sqr (A/2) where a is the single side.





drop the negative answer





a = sqr (A/2)


b = a/A


L = 2a + b|||Let x be the width and y be the length of the rectangle.


x+2y is the perimeter (because of the river).


x+2y = k -----------(1)


area = xy = A (assume)


You have to minimize xy


If you know the fence%26#039;s meaurements , solve for y in terms of x from (1)


area A= x(k-x)/2


dA/dx = 0


k is known.


once you solve for x from dA/dx =0, solve for y using x+2y=k


verify d^2A/dx^2 %26lt; 0|||It sounds like the problem is saying the area is fixed, correct? If so then the other solutions that do not have the area in the final answer are incorrect.





Let%26#039;s say x is the dimension of the field along the river, A is the given area, and p is the perimeter of the fence. In that case the other dimension of the rectangle would be A/x, and we%26#039;re trying to minimize p while holding A constant.





Because we don%26#039;t need fencing along the river, the perimeter is p = x + 2(A/x) = x + 2A*x^-1. To minimize, we take the derivative dp/dx and set it to zero; by the power rule we get dp/dx = 1 + 2A*(-1)*x^-2 = 1 - 2Ax^-2. So setting that to zero and solving, we get





0 = 1 - 2Ax^-2


2Ax^-2 = 1


2A = x^2


鈭?2A) = x





This means that the dimension along the river should be 鈭?2A) in length. To find the other dimension, we substitute into A/x and get





A/鈭?2A) = 鈭?A^2/2A) = 鈭?A/2)





So the dimension perpendicular to the river should be 鈭?A/2).