Prove that between rectangles of a given $A$ area the square is the one with lower perimeter?

Question

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?

Answer 1

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 $\sqrt{A}$.

Answer 2

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 $\sqrt{A}=a$

Answer 3

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. $a=b$.

Answer 4

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.