Finding the graph of peculiar function

Question

Area bounded by $$2<|x+3y|+|x-y|<4 $$

I tried by using |x-y|=X then I realised it will not work since the other line is not perpendicular. I know it could be done by separately calculating via piecewise definition but it will take a long time so it can't be solved that way in exam hall.

Please help me out. Thanks in advance.

Answer 1

This is a more geometrical approach to give you an idea what the graph looks like if you're in a pinch during an exam.

This inequality describes the area between two shapes. The inner shape has edges described by $|x + 3y| + |x-y| = 2$, the outer has edges described by $|x + 3y| + |x-y| = 4$.

A good start is to look at where the corners of the shape are. These are the points where one of the inequalities changes sign ($x-y = 0$ or $x + 3y = 0$).

Consider the former first. Along the line $x - y = 0, y = x$, so substitute this into the formula and obtain...

$$2 < |x + 3x| + |0| < 4 \implies 2 < |4x| < 4.$$

Note $|4x| = 2 $ when $x = \pm\frac{1}{2}$ and $|4x| = 4$ when $ = \pm 1$. Plug these values into the equation $x = y$ to get the points $(\frac{1}{2}, \frac{1}{2}), (-\frac{1}{2}, -\frac{1}{2}), (1, 1)$ and $(-1, -1)$. The first two are corners of the inner shape as they correspond to $|4x| = 2$, and the second two are corners of the outer shape as they correspond to $|4x|=4$.

Now the latter. Along the line $x + 3y = 0, x = -3y$. Substitute this into the formula to obtain...

$$2 < |0| + |-3y - y| < 4 \implies 2<|4y|<4$$

Similarly we get $y = \pm \frac{1}{2}$ and $y = \pm 1$. Plug these values into $x = -3y$ to get the points $(-\frac{3}{2}, \frac{1}{2}), (\frac{3}{2}, -\frac{1}{2}), (-3, 1)$ and $(3, -1)$. Again, the first two are corners of the inner shape and the second two are corners of the outer shape.

Now we can plot the points to see what the shapes look like! Note the points can be joined by straight lines as the equation is \textbf{linear} in $x$ and $y$.

We have two parallelograms. The area described by the inequality is the area between them, so we can find the area of the larger and subtract the area of the smaller. The area of a parallelogram is given by the width of its base multiplied by its perpendicular height, so we have...

$$A = 4 \times 2 - 2 \times 1 = 6.$$

Answer 2

Let $x+3y=a$ and $x-y=b$.

Thus, we obtain $$2<|a|+|b|<4,$$ which in coordinates $(a,b)$ gives a difference of areas of two squares: $$\frac{8\cdot8}{2}-\frac{4\cdot4}{2}=24,$$ which is $|\Delta|$ of the need area, where $\Delta$ is a determinant of the system: $$\Delta=1\cdot(-1)-1\cdot3=-4,$$ which gives that the area it's $$\frac{1}{4}\cdot24=6.$$