a farmer wants to make a rectangular paddock with an area of $ 4000 m^2$ However, fencing costs are high and she wants the paddock to have a minimum perimeter.
I have found the perimeter:
$$x\cdot y = 4000\\ y = \frac{4000}{x}$$
$$\begin{align}\text{Perimeter} &= 2x + 2y\\ &= 2x + 2(4000/x)\\ &= 2x + (8000/x)\end{align}$$
How do I find the dimensions that will give the minimum perimeter?
You could just minimize the function that you found for the perimeter. The minimum is attained when the derivative is zero. So calculate: \begin{equation} \frac{d}{dx} perimeter= \frac{d}{dx}(2x+\frac{8000}{x})=2-\frac{8000}{x^2}=0. \end{equation} This gives $2=\frac{8000}{x^2}$, so $4000=x^2$. We then get \begin{equation} x=\sqrt{4000}, \end{equation} since the negative solution is not an option.
As $x>0$
by AM GM inequality $$\dfrac{2x+\dfrac{8000}x}2\ge\sqrt{2x\cdot\dfrac{8000}x}$$
which can also be written as
$$2x+\dfrac{8000}x=\left(\sqrt{2x}-\sqrt{\dfrac{8000}x}\right)^2+2\sqrt{2x\cdot\dfrac{8000}x}\ge2\sqrt{2x\cdot\dfrac{8000}x}$$
Rectangle with maximal area and minimal perimeter is square if $x\cdot y=4000$ then $x=y$ so $x^2=4000$ and $x=\sqrt{4000}=63,2455...$
$$ 2x+2y=p,\, y= p/2 -x ; \, A= y\cdot x = (p/2 -x)\cdot x $$
$$ \frac{dA}{dx} =0 \rightarrow x = p/4 =y ,\quad A= p^2/16. $$
For maximum area it should be a square of quarter perimeter length as side.
If you use $A.M \geq G.M$, you get $2x + 2y \geq 4\sqrt{xy} = 80\sqrt{10}$ and equality holds iff $x=y$.
To minimise $2(x+y)$ subject to $xy=A$ note that: $$(x+y)^2=4xy+(x-y)^2=4A+(x-y)^2$$
Now since $x$ and $y$ are both positive, the minimum value of $2(x+y)$ occurs when $4(x+y)^2$ is a minimum and hence when $(x+y)^2$ is a minimum.
Since $(x-y)^2$ is non-negative, the minimum occurs when $x-y=0$ ie when you have a square and $x+y=2\sqrt A$ and $2(x+y)=4\sqrt A$
