The square $A B C D$ drawn in the figure below has a side of $3 \mathrm{~cm}$.

![](https://cdn.mathpix.com/cropped/2024_05_01_d02c2755ad3373bde08ag-06.jpg?height=304&width=325&top_left_y=1138&top_left_x=742)

The points $P$ and $Q$ can be moved along the segments $A B$ and $A D$ respectively such that the length of the segment $A P$ is half the length of the segment $A Q$.

a) Determine the value of the area of the hatched quadrilateral in terms of the length of the segment $A B$.

b) Determine the maximum area that the hatched quadrilateral can assume.
Solution

a) Let's call $x$ the length of the segment $A P$ and denote by $f(x)$ the value of the area of the quadrilateral $B C Q P$ as a function of $x$, measured in square centimeters. Since the length of $A P$ is half the length of the segment $A Q$, the maximum value that $x$ can assume is $3 / 2 \mathrm{~cm}$. Our goal is to find an expression for $f(x)$.

Since the segment $A Q$ is twice the length of the segment $A P$, its length is equal to $2 x$. Therefore, the segment $Q D$ measures $3-2 x$. From this, we conclude that the area of the triangle $A P Q$ is given, in square centimeters, by $(x \times 2 x) / 2 = x^{2}$ and the area of the triangle $Q D C$ by $[3(3-2 x)] / 2$.

Since the area of the shaded region is equal to the area of the square $A B C D\left(9 \mathrm{~cm}^{2}\right)$ minus the sum of the areas of the triangles $A P Q$ and $Q D C$, we have that $f(x)$ will be given, in square centimeters, by:

$$
f(x)=9-\frac{3(3-2 x)}{2}-x^{2}=-x^{2}+3 x+\frac{9}{2}
$$

for $x \in [0, 3 / 2]$.

b) The quadratic function $f(x)=-x^{2}+3 x+9 / 2$ obtained in the previous item provides the area of the quadrilateral $B C Q P$ in terms of the value of the length of the segment $A P$ denoted by $x$. The graph of this function is a downward-opening parabola, as shown in the figure below:

![](https://cdn.mathpix.com/cropped/2024_05_01_d02c2755ad3373bde08ag-26.jpg?height=370&width=685&top_left_y=381&top_left_x=567)

Thus, our goal is to find the maximum value assumed by this function with $x$ varying between 0 and $3 / 2$. For a parabola that is the graph of a function of the form $h(x)=a x^{2}+b x+c$, the $x$-coordinate of the vertex is given by $x_{v}=\frac{-b}{2 a}$. If the parabola is downward-opening, then the corresponding function achieves its maximum exactly at $x=x_{v}$. In our case, with $a=-1$ and $b=3$, we have $x_{v}=3 / 2$. Since $3 / 2$ belongs to the interval stipulated for the values that $x$ can assume, the maximum value assumed by $f$ is equal to $f(3 / 2)=27 / 4 \mathrm{~cm}^{2}$.