I have no idea why I am getting the wrong answer again.
A farmer wants to fence an area of 1.5 million square feet in a rectangular field and then divide it in half with a fence parallel to one of the sides. How can he do this so as to minimize the cost of the fence?
I know that my two functions are
$l2w = 1.5million$ where l is length and w is width which I will have two of since it is cut in half by a parallel fence.
$2l + 3w = p$ this is because there are 2 length sections and 3 width since a parallel fence adds one length.
I subsitute in
$\frac {1500000}{w} + 3w = a$
the derivative
$\frac{-1500000}{w^2} + 3 = a$
find the zero gives me nonsense and a wrong answer.
Best Answer
You only made one computational error and a notational error.
Here is your solution (at least the first part that follows) re-written:
You are minimizing the cost of the fence, so you want to minimize the perimeter of the enclosure. The perimeter $P$ is $$\tag{1} P=2l+3w. $$ where $l$ is the length and $w$ is the width of the enclosure. It is assumed the division of the original fence is made across a width (hence the $3w$).
We want to write $P$ in terms of one variable. For this we use the given information that the area is $1.5\cdot10^6\,\rm ft^2$. So $$lw= 1.5\cdot10^6\,\rm ft^2,$$ or, $$\tag{2} l={1.5\cdot10^6\over w} $$ Substituting $(2)$ into $(1)$ gives $$ P=2\bigl({1.5\cdot10^6\over w}\bigr) +3w $$ or $$ P(w)= {3\cdot10^6\over w} +3w $$ (this is where you made your mistake).
We also need to find the appropriate range of values for $w$. Here, $w$ could be possibly any positive, finite length.
So, you want to find the minimum value of $P$ over the interval $(0,\infty)$.
Towards this end, evaluate $$ P'(w)= - {3\cdot10^6\over w^2} +3. $$ Set $P'(w)=0$ and find the solutions in the interval $(0,\infty)$: $$ - {3\cdot10^6\over w^2} +3 = 0 \quad\iff\quad {3\cdot10^6 } =3w^2\quad\iff\quad w=10^3. $$
So $w=1000\,\rm ft$ gives the only critical point of $P$ in the interval $(0,\infty)$. We need to examine what goes on near the endpoints of our interval before declaring that $1000\,\rm ft$ gives the answer. As you can easily convince yourself $\lim\limits_{w\rightarrow 0^+} P(w)$ and $\lim\limits_{w\rightarrow \infty} P(w)$ are both infinite. So the cost of the enclosure is minimized when $w=1000\,\rm ft$.
And to explicitly answer the question (don't forget to do this): "he can do this by taking the width to be $1000\,\rm ft$, the length to be ${1.5\cdot 10^6\over1000}=1500\,\rm ft$ and spliting the enclosure by constructing another width".