Area Calculation with Absolute Value Inequality $|x+2y|+|3x+4y|\le 5$

Question

I am trying to evaluate the area of a plane region defined by the inequality $|x+2y|+|3x+4y|\le 5$. I have plotted the graph of this inequality using an online tool (see image above), but I am unsure how to proceed with calculating the area.

I am hoping someone can provide me with a method to evaluate the area of this plane region, whether it be through geometric or algebraic techniques. Any help or insight would be greatly appreciated. Thank you in advance.

Answer 1

Hint.

Making $u = x+2y,\ v = 3x+4y$ we have $|u|+|v|\le 5$ and in coordinates $(u,v)$ the area is 50. Introducing the area correction due to the coordinates transformation we have

$$ \left(\matrix{du\\ dv }\right)=\left(\matrix{1 & 2\\ 3 & 4}\right)\left(\matrix{dx\\ dy }\right) $$

and $\left|\det\left(\matrix{1 & 2\\ 3 & 4}\right)\right| = 2$ hence

$$ \text{area}\left(|x+2y|+|3x+4y|\le 5\right) = \frac{50}{2}=25 $$

Answer 2

Compute base $\times$ height.

Length of base is $[-10-(-5)]$ or $[-5-(-10)]$, so
$base=15$.

Length of height is (f.e. at $x=0$) obtained via

$|(0+2y)|+|(0+4y)|=5$
$(0+2y)+(0+4y)=10$ or $y=\frac{5}{3}$, so
$height=\frac{5}{3}$.

The area is then $15\times\frac{5}{3}=\frac{75}{3}=25$.