Optimization of a Farmer's Fence
Sebastian Wright 
A farmer wants to fence an area of 24 million square feet in a rectangular field and then divide it in half with a fence parallel to one of the sides of the rectangle. What should the lengths of the sides of the rectangular field be so as to minimize the cost of the fence?
I need to find the:
smaller value:
larger value:
this was my process.
a = xy | 2.4 x 10^6 = xy | p = 3x + 2y | x is the length | y is the width |
I isolated y from the area formula, so I got this y = (2.4 x 10^6)/x
and then I substituted it into the "p" function
to get this p = 3x + 2(2.4 x 10^6 / x) => p = 3x + 4.8 x 10^6
then I found the critical points by taking the derivative of p.
which came out to be
0 =-4.8 x 10^6/x^2 + 3
x = 1264.911064 which is the smaller side.
I would get the larger side by plugging 1264.911064 into p = 3x + 2(2.4 x 10^6 / x).
But it seems my answer is wrong when I enter it into webassign.
$\endgroup$ 13 Answers
$\begingroup$This can be done easily using calculus.
We know that there will be 3 widths and 2 lenghts and that the area is 24'000'000. So $(Lenght)(Width)=24'000'000$ and we want to minimize the fencing so minimize $3w+2l=F$ which can be written as $3(\frac{24'000'000}{l})+2l=F$ since we can write the width as: $\frac{24'000'000}{l}=w$.
Calculate the derivative of $F$ to get $F'= 2- \frac{72'000'000}{l^2}$ and is eqaul to $0$ when $l=6000$ and $w=4$
You can use this to calculate the fencing needed.
$\endgroup$ $\begingroup$The length of the fence is $2a+3b$. The area is $A=ab$. Thus $a=A/b$, and the length is $2A/b+3b$. Setting the derivative with respect to $b$ to $0$ yields
$$ -\frac{2A}{b^2}+3=0 $$
and thus
$$ b=\sqrt{\frac{2A}3}\;. $$
$\endgroup$ $\begingroup$Optimization requires first finding a function we want to optimize (usually max or min and usually for calc I two or less variables), then using some known relationship of the variables to turn the optimizable function into a one variable equation.
Here we have $$A=xy$$ and $$P=2x+3y$$ (the 2x is two sides and the 3y comes from having two parallel sides and a third splitting the enclosure)
We \emph{know} the area is $A=24=xy$, so we can solve for $x$ as $x=\frac{24}{y}$, then substitute it into the fencing equation we want to maximize (P = "perimeter"). So we get $$P=2x+3y = 2(\frac{24}{y})+3y$$ and optimize by taking the derivative to get $$P'(y)=3-\frac{48}{y^2}=0$$. We solve for $y$ as follows:
$0 = 3-\frac{48}{y^2}$
$0 = 3y^2-48$
$y^2=16$
$y=4$ since a distance must be non-negative.
We can substitute to get $A=24=xy=4x$ so $x=6$.
Thus the amount of fencing is $P=2x+3y=2(6)+3(4)=24$.
{\color{blue}An interesting and better question to ask yourself is why do we always get that the total amount of horizontal fencing is equal to the total amount of vertical fencing in these optimization questions? Ex: If you do one of the classic build a fence along a river so we only need to fence 3 sides, what is the max area with 100ft of fence? We will surely get a 25 x 50 x 25 x river setup, splitting the total distances vertically and horizontally into 50 and 50. For calc I questions this will always be the case and try to reason why.}
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