Consider in the Cartesian coordinate system those points for which there exists a rectangle such that the first coordinate of the point is the perimeter of the rectangle, and the second coordinate is the area of the same rectangle. Characterize the domain defined by these points.
A point $P(k ; t)$ belongs to the given domain if and only if the system of equations

$$
\begin{aligned}
k & =2(x+y) \\
t & =x y
\end{aligned}
$$

has a solution in positive numbers, where $x, y$ represent the sides of a rectangle. Thus, $k>0, t>0$ is a necessary condition. Based on the relationship between the roots and coefficients, this is true if and only if $x$ and $y$ are the roots of the equation

$$
u^{2}-\frac{k}{2} u+t=0
$$

Since two numbers are positive if and only if their sum and product are positive, if $k>0, t>0$ and equation (2) is solvable, then the roots are necessarily positive.

Thus, we have that the point $P(k ; t)$ belongs to the given domain if and only if $k>0, t>0$ and equation (2) has a solution, which is

$$
\frac{k^{2}}{4} \geq 4 t
$$

Based on this, the domain in the rectangular coordinate system defined by the $k$ and $t$ axes is the part of the first quadrant bounded by the $k$ axis and the parabola

$$
t=\frac{k^{2}}{16}=\left(\frac{k}{4}\right)^{2}
$$

The points on the parabola belong to the domain, while the points on the $k$ axis do not.

![](https://cdn.mathpix.com/cropped/2024_05_02_20a55503dab64940e9c1g-1.jpg?height=529&width=1006&top_left_y=1134&top_left_x=554)

Remarks. For points on the parabola, $\frac{k^{2}}{4}=4 t$; thus, the roots of equation (2) are equal, indicating that these points represent squares.

2. The above question can also be generalized. The consequence of the statement in Exercise 1654 (1977, No. 2, p. 73) is that the domain does not change if we inscribe a quadrilateral instead of a rectangle. For triangles, we get a narrower domain than the one above. If we inscribe an arbitrary plane figure instead of a rectangle, the answer can be given using the isoperimetric inequality for the perimeter $k$ and area $t$ of the arbitrary plane figure:

$$
\frac{k^{2}}{4 \pi} \geq t
$$

This inequality expresses that among plane figures with a given perimeter, the circle has the largest area.