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

Just solve $lb = 2l+2b$ for $b$, giving $b = \frac{2l}{l-2}$. There are infinitely many solutions. — rogerl, May 12 '14 at 17:01
@Awesome: Certainly, one could/should ask about a rectangle for which the number of units in the perimeter is equal to the number of square-units in the area. That description is quite a mouthful, though, so it's not-unreasonable to phrase the condition as "perimeter equals area". — Blue, May 12 '14 at 17:11
Assuming we measure area in square of the unit we measured length with and are comparing magnitudes only, this question makes sense. — evil999man, May 12 '14 at 17:16
@RavindraSahay You deserve a downvote. http://lmgtfy.com/?q=area+perimeter+equal+rectangle First hit shows solution. This question does not show any research effort. It says that on downvote icon.:D — evil999man, May 12 '14 at 17:21
I used hit and trail method (clearly mentioned) finding all possible combinations to get the answer. trying 1 with 2, then 3 and so on up-to 20. and also i tried solving it in form of an equation. Imagine forming combinations. @Awesome — Ravindra Sahay, May 12 '14 at 17:26

score 1 · Answer 1 · answered May 12 '14 at 17:07

1

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)

answered May 12 '14 at 17:07

abstract

199

hmm. coordinate geometry I am not very good at it but still i am getting in something but it would be more respected that you show the method that you followed to get to this point – Ravindra Sahay May 12 '14 at 17:10
@RavindraSahay I see no coordinate geometry. – evil999man May 12 '14 at 17:17
and what are quadrants my friend and all equation can be plot on coordinate system. @Awesome – Ravindra Sahay May 12 '14 at 17:27
download a graph plotting software, plot the function y=2x/(x-2), and you will see a graph, the ordered pairs in first quadrant contains all integral values for (b,l), and non-integral as well, as the results of von Eitzen suggest. – abstract May 12 '14 at 17:31
just google 2x/(x-2) – evil999man May 13 '14 at 10:54

score 0 · Answer 2 · answered May 12 '14 at 17:12

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 $2$.

score 0 · Accepted Answer · answered May 12 '14 at 17:13

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.

A mental question

3 Answers3