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The area of the region of the plane bounded by $\max(|x|,|y|) \leq 1$ and $xy \leq 1/2$

Can you explain, in simple terms how the term $\max(|x|,|y|) \leq 1$ represents a square?

The answers below did not give much guidance.

Area bounded by $\max (|x|, |y|)\leq 1$ and $xy \leq \frac{1}{2}$

Show that the $\max{ \{ x,y \} }= \frac{x+y+|x-y|}{2}$.

Moreover, I also do not understand how can that max term be represented as a function purely in $x$ to perform normal integration thereafter. There is another answer that uses double integrals. Is there a simpler method? In some solution, the square at the origin is rotated to look like a diamond.

Michael Hardy
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mathx
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2 Answers2

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In order to understand how $\max(|x|,|y|)$ represents a square, consider its complement region. The graph cannot exist in the regions $|x|>1$ and $|y|>1$. Hence the only possibility where it may exist are the intervals: $$-1\leq x\leq 1$$ and $$-1\leq y \leq 1$$ In these intervals, either $|x|$ or $|y|=1$, hence we must have the perimeter of the $1×1$ square as the graph, as shown in the linked answer. If you are having trouble understanding the double integration, you may completely skip it. The hyperbola $y=\frac {1}{2x}$ intersects the square at points $(\frac 12, 1)$ and $(1,\frac 12)$. Thus, the area that we need can be represented as: $$A=B-\int_{\frac 12}^1 \frac {dx}{2x}$$ where $B$ is area bounded by $x$-axis, $y=1$, $x=\frac 12$ and $x=1$. Thus, $B=\frac 12×1=\frac 12$. Thus $A$ can be easily evaluated.

Michael Hardy
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Ritam_Dasgupta
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  • Thanks for that concise explanation. Nothing more than what was needed. – mathx Jul 14 '21 at 15:13
  • Is this a square and not a diamond? – mathx Jul 14 '21 at 15:14
  • I changed "compliment" to "complement". Those are two different words. If I say your answer is brilliant, that's a compliment (with an "i"). But here you're writing about a complement (with an "e"). – Michael Hardy Jul 14 '21 at 15:16
  • @MichaelHardy If you know either of the two, you know what the difference is. But still, you got a chance to add $e^1$ somewhere. – mathx Jul 14 '21 at 15:19
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    @Michael Hardy I am a phone user, that was auto-correct. I am aware of the difference between the two words. – Ritam_Dasgupta Jul 14 '21 at 15:29
  • @mathx, the expression $max(|x|,|y|)=1$ does not represent a diamond, but a square with edges along the coordinate axes. Think about it, $(1,1)$ is certainly a point on the curve, yet it wouldn't have been included, had we taken a diamond. – Ritam_Dasgupta Jul 14 '21 at 15:30
  • @Ritam_Dasgupta Correct. Again, just what was needed. Thanks – mathx Jul 14 '21 at 15:48
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To say that $\max\{|x|,|y|\} \le 1$ is the same as saying that both $|x|\le1$ and $|y|\le 1.$

$|x|\le1$ is equivaent to $-1\le x\le+1.$

So $-1\le x\le+1$ and $-1\le y\le+1.$

That is a square.

Michael Hardy
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