13. The living room of Filippo's house has a rectangular floor plan. Filippo noticed that if he plugs the vacuum cleaner into the outlet near the entrance door, he can clean the entire floor: this means that all points on the floor are less than $5 \mathrm{~m}$ away from the point on the wall where the outlet is located. What is the maximum area of the living room (in $\mathrm{m}^{2}$)?
13. The answer is 25. Consider the rectangle that forms the floor plan of the living room, and orient it so that the outlet $P$ is on the horizontal side $A B$, dividing it into two segments of length $a$ and $b$. We also call $c$ the length of the two vertical sides $B C$ and $D A$, as shown in the figure. In this way, the area of the living room is $a c + b c$.

![](https://cdn.mathpix.com/cropped/2024_04_17_c5c03d45c82457ddb3cag-17.jpg?height=317&width=468&top_left_y=144&top_left_x=840)

For vertices $C$ and $D$ to be at a distance of 5 or less from $P$, we must have $\sqrt{b^{2} + c^{2}} \leq 5$ and $\sqrt{a^{2} + c^{2}} \leq 5$, respectively. By a known inequality, we have $a c \leq \frac{a^{2} + c^{2}}{2}$, with equality when $a = c$, and similarly $b c \leq \frac{b^{2} + c^{2}}{2}$. Combining these inequalities, we get that the area of $A B C D$ is

$$
a c + b c \leq \frac{a^{2} + c^{2}}{2} + \frac{b^{2} + c^{2}}{2} \leq \frac{5^{2}}{2} + \frac{5^{2}}{2} = 25.
$$

To verify that this is indeed the maximum area, we want to construct a configuration in which all these inequalities become equalities. We want, in particular, to have $a = c$ and $25 = a^{2} + c^{2} = 2 a^{2}$, and solving gives $a = c = \frac{5}{\sqrt{2}}$. Similarly, we get $b = c = \frac{5}{\sqrt{2}}$. Indeed, choosing $a = b = c = \frac{5}{\sqrt{2}}$, all points inside the rectangle are at most $\frac{5}{\sqrt{2}}$ away from $P$ along both the horizontal and vertical axes, so their distance from it is at most

$$
\sqrt{\left(\frac{5}{\sqrt{2}}\right)^{2} + \left(\frac{5}{\sqrt{2}}\right)^{2}} = \sqrt{25} = 5.
$$

Second solution. Consider the points $C'$ and $D'$ that are obtained by reflecting $C$ and $D$ across the side $A B$.

![](https://cdn.mathpix.com/cropped/2024_04_17_c5c03d45c82457ddb3cag-17.jpg?height=320&width=320&top_left_y=1479&top_left_x=914)

Then the rectangle $C D D' C'$ (which has double the area of $A B C D$) lies within the circle centered at $P$ with radius 5. Therefore, finding the maximum area of $A B C D$ is equivalent to constructing the rectangle of maximum area that fits inside a circle, and it is known that this is the square. We then have $\overline{C D} = \overline{D D'} = \overline{D' C'} = \overline{C' C}$, and we conclude that $A D$ and $B C$ must be half the length of $C D$.