# Inequalities and area

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 $x\geq-2y$

$$\left|x-2y\right|+\left|x+2y\right|=4y=8$$

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

$$\left|x-2y\right|+\left|x+2y\right|=-2x=8$$

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

$$\left|x-2y\right|+\left|x+2y\right|=-4y=8$$

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:

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 $2(6-2\log 4)$.