## Task A-2.8. (20 points)

Determine $a>0$ such that the area of the figure bounded by the graphs of the functions

$$
y=|a x-3|+|a x+3| \quad \text{and} \quad y=10
$$

is equal to 8.
## Solution.

The function $y=|a x-3|+|a x+3|$ can be written piecewise without absolute values:

$$
y=\left\{\begin{array}{cl}
-2 a x, & x \in\left(-\infty,-\frac{3}{a}\right) \\
6, & x \in\left[-\frac{3}{a}, \frac{3}{a}\right] \\
2 a x, & x \in\left(\frac{3}{a}, \infty\right)
\end{array}\right.
$$

Let's sketch the graphical representation of the observed functions and the desired area:

![](https://cdn.mathpix.com/cropped/2024_05_30_59925bf190fc107f19b7g-10.jpg?height=816&width=619&top_left_y=1094&top_left_x=730)

The region bounded by the graphs of the observed functions is a trapezoid with vertices $A\left(-\frac{3}{a}, 6\right), B\left(\frac{3}{a}, 6\right), C\left(\frac{5}{a}, 10\right), D\left(-\frac{5}{a}, 10\right)$.

Its area is $P=\frac{1}{2} \cdot\left(\frac{6}{a}+\frac{10}{a}\right) \cdot 4=\frac{32}{a}$.

The area of the trapezoid is 8 if $a=4$.

## DISTRICT/SCHOOL COMPETITION IN MATHEMATICS

## 3rd grade - high school - A category

29. January 2009.