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?
On
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
$$2x+\dfrac{8000}x=\left(\sqrt{2x}-\sqrt{\dfrac{8000}x}\right)^2+2\sqrt{2x\cdot\dfrac{8000}x}\ge2\sqrt{2x\cdot\dfrac{8000}x}$$
On
Rectangle with maximal area and minimal perimeter is square if $x\cdot y=4000$ then $x=y$ so $x^2=4000$ and $x=\sqrt{4000}=63,2455...$
On
$$ 2x+2y=p,\, y= p/2 -x ; \, A= y\cdot x = (p/2 -x)\cdot x $$
$$ \frac{dA}{dx} =0 \rightarrow x = p/4 =y ,\quad A= p^2/16. $$
For maximum area it should be a square of quarter perimeter length as side.
On
If you use $A.M \geq G.M$, you get $2x + 2y \geq 4\sqrt{xy} = 80\sqrt{10}$ and equality holds iff $x=y$.
On
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 $2(x+y)=4\sqrt A$
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.