Find the area of the region described by $|5x|+|6y| \le 30 $ 
Find the area of the region described by $|5x|+|6y| \le 30 $ (where
   $|z|$ denotes the absolute value of $z$).

My effort
Imagining a number line and interpreting the problem as the request to have the sum of the distances of $5x,6y$ from $0$ to be less or equal to $30$ ,we have that  :
i) for $|5x| \le 30-h $, where $0 \le h \le 15 $, then $ |6y| \le h$ (and also the other way around : $ |6y| \le 30-h $ and $|5x| \le h$).
Therefore $15\le 5x \le 30 $ and $-30 \ge 5x \ge -15 $ which implies $x \in [-6,-3]\cap [3,6] $ and so we also we find $y \in [-15/6,15/6] $.
Inverting the cases we find $x \in [3,-3]$ and $y\in [-5,-15/6]\cap [15/6,5]$.
By plotting several points for $ 0 \le x \le 6 $ and using the fact that we have symmetry about the origin $(0,0)$ we find that the requested area is a square with side $\sqrt{61}$ (since one side will have coordinates $(-6,0)$ and $(0,-5)$)

Question
Is my effort correct ?How could I have simplified it (I think I have
  made the problem harder than it needed to be in some part) ? Are there
  other (easier ?) ways to tackle the problem ?

 A: 
$$|5x| + |6y| \le 30$$

There are a couple ways to approach this
Approach 1:
Consider only the $x \ge 0$ and $y \ge 0$ quadrant.  The shape enclosed is a triangle, with $x$-intersept at $5\cdot 0 + 6y = 30$, that is $y=5$.  Similarly the $y$-intersept is at $x = 6$.  So the area of the triangle is $\frac{1}{2}5\times 6 = 15$, and the area across all 4 quadrants is then $4\times 15 = 60$.
Approach 2:
$|x| + |y| \le 1$ is a diamond with edgelength $\sqrt 2$.  Thus the area of the diamond is $2$.  $|x| + |y| \le 30$ increases the radius of the diamond $30$ times so the area is $2\times 30 \times 30 = 1800$.  $|5x| + |y| \le 30$ decreases all $x$ values by a factor of $5$ so the area is $1800 / 5 = 360$.  $|5x| + |6y| \le 30$ decreases all $y$ values by a factor of $6$ so the area is $360 / 6 = 60$.
A: Your effort is correct, but the region is a diamond shape, not a square.
Another approach is to view the inequality $|5x| + |6y| \le 30$ as a region in the $(x,y)$ plane. Notice that if $(x,y)$ satisfies the inequality, then so do $(-x,y)$, $(x,-y)$, and $(-x,-y)$. This means that the region is symmetric about the $x$-axis as well as the $y$-axis. So it is enough to picture the region in the first quadrant, i.e. consider just the case where $x>0$ and $y>0$; the area of the region is then four times the area in the first quadrant.
The region in the first quadrant is a triangle with base $6$ and height $5$: it's bounded by the line $5x + 6y = 30$ in the first quadrant, which intersects the axes at $(6,0)$ and $(0,5)$. So the area of the overall region is $4 \cdot (\frac12 \cdot 6 \cdot 5) = 60$.
A: Hint:
The best general way is to separate the cases of the absolute values and find the regions delimited by the single case ( the final region is the union of such regions). So we have:
$$
\begin{cases}
x\ge 0 \quad \land \quad  y \ge 0\\
5x+6y\le 30 \rightarrow y\le 5-\frac{5}{6}x
\end{cases}
$$
$$
\begin{cases}
x< 0 \quad \land \quad  y \ge 0\\
-5x+6y\le 30 \rightarrow y\le 5+\frac{5}{6}x
\end{cases}
$$
$$
\begin{cases}
x< 0 \quad \land \quad  y < 0\\
-5x-6y\le 30 \rightarrow y\ge -5-\frac{5}{6}x
\end{cases}
$$
$$
\begin{cases}
x \ge 0 \quad \land \quad  y < 0\\
5x-6y\le 30 \rightarrow y\ge -5+\frac{5}{6}x
\end{cases}
$$
Now represent this regions and you can easily find that the union is a parallelogram with vertex at $(0,5),(6,0),(0,-5),(-6,0)$ that has orthogonal diagonals so you can easily find its area.
In this case, noting the symmetry with respect to the $x$ and $y$ axis, it is sufficient to solve only one of this systems.
