A mental question

Question

We are given a rectangle whose perimeter and area are equal. We have to find their length and breadth.(Both length and breadth should differ). I solved the question via hit and trial method and got the answer 6&3, but I want to know if the question can be done as an equation. Because when I try, l*b=2l+2b ~(l*b)/2 = l+b ~After this I get stuck and one time i got on and found length and breadth as -2 and 1 which practically is not possible as length and/or breadth cannot be negative

Answer 1

If you want integer solutions, there are only few: From $$b=\frac{2l}{l-2}$$ we see that $l-2$ is a positive divisor of $2l$, especially $l\ge 3$. As $\gcd(l,l-2)=\gcd(l,2)\le 2$, we even have that $l-2$ divides $4$. Thus leaves us with

• $l-2=1$ and then $l=3$, $b=6$
• $l-2=2$ and then $l=4$, $b=4$, which is not allowed
• $l-2=4$ and then $l=6$, $b=3$, the "dual" to the first

All other solutions (with arbitrary $l>2$, $l\ne 4$) are valid, but not integer solutions.

Answer 2

Plot length(l) along Y axis, breadth(b) along X axis; ordered pairs (b,l) satisfying the relation y=2x/(x-2), and lying in the first quadrant is the set of all positive-real-values of (b,l)

Answer 3

We have $b=2a/(a-2)$. In case you're looking for integer solutions, $a$ must be at least $3$. You'll find integer solutions for $a\in\{3,4,6\}$. These are the only integer solutions: if $a=7$, we have $b=2.8$. Now $a\mapsto2a/(a-2))$ is strictly decreasing for $a>2$ and it's limit for $a\to\infty$ is $2$.