Find the area of the region which contains all the points satisfying the inequalities $|x-2y|+ |x+2y|\leq 8$ and $xy\geq 2$.

My attempt- I considered the case when $x>2y$.the second inequality gives $2y^2\geq 2$,which gives $y^2\geq 1$. And first inequality reduces to $2x\leq 8$, $x\leq 4$. After all this I could not proceed and am far behind the numerical answer of $2(6-2\ln 4)$ sq. units.


Well, let's first think of the boundary $\left|x-2y\right|+\left|x+2y\right|=8$.

Case 1: $x\geq2y$ and $x\geq-2y$

$$\left|x-2y\right|+\left|x+2y\right|=2x=8$$ Case 2: $x\leq2y$ and $x\geq-2y$


Case 3: $x\leq2y$ and $x\leq-2y$


Case 4: $x\geq2y$ and $x\leq-2y$


Piecing it all together, we have the lines $x=4$, $x=-4$, $y=2$, $y=-2$, which are cut into line segments by the lines $2y=x$ and $2y=-x$. These line segments come together to form a $8\times4$ rectangle centered at the origin. The region $\left|x-2y\right|+\left|x+2y\right|\leq8$ is the interior of this rectangle.

So our graph looks something like this:

enter image description here

We need to find the intersection of the blue and red areas. We can just find the area of the upper right intersection, and then multiply the result by $2$. Note that the blue and red graphs intersect in the first quadrant at $(1,2)$ and $\left(4,\frac{1}{2}\right)$. Consider a rectangle with vertices $(1,2)$, $(1,0)$, $(4,0)$, and $(4,2)$. Note that it has an area of 6. Note that the area of the aforementioned rectangle minus the upper left intersection area is equal to $\int_1^4 \frac{2dx}{x}=2\ln(4)-2\ln(1)=\ln(4)$. So the area of one intersection is $6-2\ln(4)$, and our final answer is $2(6-2\ln(4))=12-4\ln(4)$.


Alternative approach.

We change the variables: let $u=x-2y$ and $v=x+2y$. Then $x=(u+v)/2$, $y=(v-u)/4$. Moreover the domain $$D=\{(x,y): |x-2y|+ |x+2y|\leq 8, xy\geq 2\}$$ is trasformed into $$D'=\{(u,v): |u|+ |v|\leq 8, v^2-u^2\geq 16\}$$ with a factor of transformation $|\partial(x,y)/\partial(u,v)|=1/4$.

Hence the area is $$\iint_D dx dy=\frac{1}{4}\iint_{D'}dudv=\frac{4}{4}\int_{u=0}^3\int_{v=\sqrt{16+u^2}}^{8-u}dv du\\=\int_{u=0}^3\left(8-u-4\sqrt{1+(u/4)^2}\right)du=12-4\ln(4).$$


We divide the region into four cases, namely

(i) $x-2y\ge0$, $x+2y\ge0$, (ii) $x-2y\ge0$, $x+2y\le0$, (iii) $x-2y\le0$, $x+2y\ge0$ and (iv) $x-2y\le0$, $x+2y\le0$.

For (i), the region is $x-2y+x+2y=2x\le8$, viz. $x\le4$, $x-2y\ge0$, $x+2y\ge0$ and $xy\ge2$. For (iii), the region is $x+2y-x+2y=4y\le8$, viz. $y\le2$, $x-2y\le0$, $x+2y\ge0$ and $xy\ge2$. Thus (i) and (iii) combined form the region $\{(x,y):1\le x\le4,2/x\le y\le2\}$, which has an area of $$ \int^4_1\left(2-\frac{2}{x}\right)dx=6-2\log 4. $$ By symmetry, the regions (ii) and (iv) combined has an area of $6-2\log 4$ also. Hence the numerical answer of $2(6-2\log 4)$.


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