Proving logical equivalence of two statements So for extra credit, our teacher told us to prove the following 2 statements are logically equivalent:


*

*The square is the quadrilateral that maximizes the area given a fixed perimeter $P$.

*The square is the quadrilateral that minimizes the perimeter given a fixed area $A$. 


My question is, $\textbf{how would I go about proving this?}$
Here is the proof I offered:
Statement 1 can be reworded as the following: 


*

*The square is the quadrilateral that maximizes the number $Y=\frac{A}{P}$


Where $A$ is the area and $P$ is the perimeter of the quadrilateral. Similarly, Statement 2 can be reworded as the following:


*The square is the quadrilateral that minimizes the number $Z=\frac{1}{Y}=\frac{P}{A}$


This means that when Y is maximized, $\frac{1}{Y}$ must be minimized. Let $S$ represent the statement "Quadrilateral S is a square of length $m$ and width $n$." Let $y$ represent "$Y$ is maximized and $z$ represent "$Z$ is minimized." Then statements 1 and 2 are equal to:


*

*$S \rightarrow y$

*$S \rightarrow z$


If we show that $y \equiv z$, then statement 1 is equivalent to statement 2. But then if $Y>\frac{mn}{2m+2n}$ for all $m$ and $n$ such that $m\neq n$, it follows that $\frac{1}{Y}<\frac{2m+2n}{mn}$ for all $m$ and $n$ such that $m\neq n$. Since statement $y$ and statement $z$ imply each other, it follows that statement 1 and statement 2 are logically equivalent. 
But of course I get it wrong, since when I reworded the statements I forgot that the area in statement 1 is a variable while in statement 2 it is a constant, and vice versa for the perimeter. I'm kind of stuck at the moment, so any help would be appreciated. 
 A: You can show that two statements, $S$ and $T$, are logically equivalent by proving that $S \Rightarrow T$ and $T \Rightarrow S$.
I'll show one direction, and you should be able to figure out the other:
Given "The square is the quadrilateral that maximizes the area given a fixed perimeter $P$.", prove "The square is the quadrilateral that minimizes the perimeter given a fixed area $A$."
Consider all quadrilaterals with the same fixed area $A$. There should be one (or more) quadrilaterals with the minimum perimeter. Take any one such quadrilateral, call it $Q$.
If $Q$ is NOT a square, then by the given statement, we can construct a square $Q^\prime$ which has the same perimeter as $Q$, but with larger area $A^\prime > A$. Furthermore, we can simply "shrink" $Q^\prime$ by an appropriate scaling factor to get a square, $Q^{\prime\prime}$ with area $A$. Since $Q^{\prime\prime}$ is smaller than $Q^\prime$, it must have a smaller perimeter. We have reached a contradiction, so our initial assumption that $Q$ is not a square is false.
A: I will hint a way to the first one and leave the rest to you:



*

*Perimeter $C=2A+2B$ (whatever the lengths of $A$, $B$, Right?)

*Area $D=A*B$ (whatever the lengths of $A$, $B$, Right?)


Now let's translate the problem to a $MAX, MIN$ problem, we can say in other words :
Prove that $Max(A*B)$ is when $A=B$, knowing that $A+B=Constant=C$ ?
OK? let's tackle this one now :
$A+B=C$ ==> $B=C-A$
$D=A*B$ 
$D=A(C-A)$
$D=-A^2+CA$ (we want to compute the maximum of $D$ with $A$ as a variable)
I think the easiest way is to see its first derivative
$D'=-2A+C$:
$D'=0$==> $A=\frac{C}{2}$ (Note that this is the maximum of $D$)
$A=\frac{C}{2}$ ==> $B=C-A=C-\frac{C}{2}=\frac{C}{2}$ (Great!)
So we proved that: in order the area of quadrilateral to be the max, the sides should be equal, knowing that the perimeter is constant.
I leave the other part as an exercise.
