Prove that between rectangles of a given $A$ area the square is the one with lower perimeter? Prove that between rectangles of a given $A$ area the square is the one with lower perimeter?
Im lost, cant even figure out what to do, or where to start?
 A: Note that if the sides are $a$ and $b$ with $a \ge b$ we have: $$(a+b)^2=4ab+(a-b)^2$$
Here $(a+b)^2$ is the square of half the perimeter, and $ab$ is fixed. We make $a+b$ as small as possible by making the positive term $(a-b)^2$ as small as possible i.e. $a=b$.
A: A rectangle of height $a$ and width $b$ as area $A = a \cdot b$ and Perimeter $P = 2 \cdot(a +b).$
Now we fix $A$ so $b = \frac{A}{a}$ and $P = 2 \cdot \left(a + \frac{A}{a} \right)$
It should be obvious that by making $a$ tiny $b$ becomes very large and so does the perimeter.
there is a minimum perimeter when $\frac{d}{da}P = 0$
$$
\frac{d}{da} P = \frac{d}{da} 2 \cdot \left(a + \frac{A}{a} \right) = 2 - \frac{2 \cdot A}{a^2} 
$$ 
Setting this to zero we have 
$$
0 = 2 - \frac{2 \cdot A}{a^2} \Rightarrow a^2 = A
$$
So you will have minimum perimeter when $a = b = \sqrt{A}$.  Which is what we were asked to prove.
A: Let $l.b$ be the length and breadth of the rectangle respectively.So perimeter of the rectangle is given by:
$$2(l+b)$$
So when is (l+b) minimum ?
Arithmetic Mean $\ge$ Geometric Mean
$$\frac{(l+b)}{2} \ge \sqrt{lb}$$
$$\frac{(l+b)}{2} \ge \sqrt{A}$$
$$2(l+b) \ge 4\sqrt{A}$$
Perimeter $\ge 4\sqrt{A}$. To get the minimum possible perimeter, $l=b=a$, and $\sqrt{A}=a$
A: We have $ab=A$ and $2a+2a=p$, $p$ is the perimeter. Then, as $A$ is fixed, $b=a/A$ and
$$
p=2(a+A/a)\ge 4\sqrt{A}.
$$
The $4\sqrt{A}$ is the perimeter of the square with side $\sqrt{A}$.
