A point $M$ describes the arc between the two vertices of an ellipse. The length of the semi-major axis is $a$ and the length of the semi-minor axis is $b$.

Let's examine the variations of the quantities $xy$ and $x+y$, where $x$ and $y$ represent the distances of point $M$ from the minor and major axes, respectively.
I. Let us assume that

$$
Y=x+y
$$

There is the following relationship between the coordinates of a point on the ellipse;

$$
\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}-1=0
$$

Let's solve for $y$ from (2) and substitute it into (1); then we get

$$
Y=x+\frac{b}{a} \sqrt{a^{2}-x^{2}}
$$

The question is, what happens to $Y$ when the value of $x$ increases from $x_{0}$ to $x_{1}$?

Let $Y_{0}$ and $Y_{1}$ be the values of $Y$ when $x=x_{0}$ and $x=x_{1}$; then the value of the expression $Y_{1}-Y_{0}$ is as follows:

$$
x_{1}-x_{0}+\frac{b}{a}\left(\sqrt{a^{2}-x_{1}^{2}}-\sqrt{a^{2}-x_{0}^{2}}\right)
$$

or

$$
x_{1}-x_{0}-\frac{b}{a} \frac{x_{1}^{2}-x_{0}^{2}}{\sqrt{a^{2}-x_{1}^{2}}+\sqrt{a^{2}-x_{0}^{2}}}
$$

which difference can also be written in the following form

$$
\left(x_{1}-x_{0}\right)\left(1-\frac{b}{a} \frac{x_{1}+x_{0}}{\sqrt{a^{2}-x_{1}^{2}}+\sqrt{a^{2}-x_{0}^{2}}}\right)
$$

The difference $Y_{1}-Y_{0}$ is positive, zero, or negative, according to whether the ratio

$$
\frac{Y_{1}-Y_{0}}{x_{1}-x_{0}}
$$

is positive, zero, or negative. But the latter is

$$
1-\frac{b}{a} \frac{x_{1}+x_{0}}{\sqrt{a^{2}-x_{1}^{2}}+\sqrt{a^{2}-x_{0}^{2}}}
$$

regardless of how small the difference $x_{1}-x_{0}$ is, even at zero. If, therefore, $x_{1}=x_{0}=x$ then

$$
Y^{\prime}=\lim \frac{Y_{1}-Y_{0}}{x_{1}-x_{0}}=\frac{a \sqrt{a^{2}-x^{2}}-b x}{a \sqrt{a^{2}-x^{2}}}
$$

or

$$
Y^{\prime}=\frac{a^{4}-\left(a^{2}+b^{2}\right) x^{2}}{a \sqrt{a^{2}-x^{2}}\left(a \sqrt{a^{2}-x^{2}}+b x\right)}
$$

Since $x$ varies between $0$ and $a$, $\sqrt{a^{2}-x^{2}}$ is always real and the denominator of $Y^{\prime}$ is always positive for positive values of the root. Its sign, therefore, depends solely on the numerator. The numerator is greater, equal, or less than zero, according to whether

$$
x \lesseqgtr \frac{a^{2}}{\sqrt{a^{2}+b^{2}}}
$$

Therefore, the table of variations of $Y$ is as follows:

$$
\frac{x \left\lvert\, o \frac{a^{2}}{\sqrt{a^{2}+b^{2}}} a\right.}{\frac{Y^{\prime} \mid+o o}{Y \mid b \sqrt{a^{2}+b^{2}}} a}
$$

that is, if $x$ increases from $0$ to $\frac{a^{2}}{\sqrt{a^{2}+b^{2}}}$, $Y$ increases from $b$ to $\sqrt{a^{2}+b^{2}}$, while if $x$ further increases from $\frac{a^{2}}{\sqrt{a^{2}+b^{2}}}$ to $a$, $Y$ decreases from $\sqrt{a^{2}+b^{2}}$ to $a$, or $Y$ has a maximum of $\sqrt{a^{2}+b^{2}}$.

The point $M$ where $Y$ has its maximum value is constructed as follows. Let the endpoints of the major axis of the ellipse be $A$ and $A^{\prime}$, and the endpoints of the minor axis be $B$ and $B^{\prime}$. Then

$$
x_{m}=\frac{a^{2}}{\sqrt{a^{2}+b^{2}}}=\frac{a^{2}}{A B}
$$

We lay off the length $O D=A B$ on the line $O B$ and connect $D$ to $A^{\prime}$. Raise a perpendicular to $A^{\prime} D$ at $A^{\prime}$ until it intersects $O B^{\prime}$ at $E$. $O E$ is the desired $x_{m}$ distance.

Indeed, from the right triangle $D A^{\prime} E$ it follows that

$$
A^{\prime} O^{2}=O D \cdot O E
$$

or

$$
O E=\frac{A^{\prime} O^{2}}{O D}=\frac{a^{2}}{A B}
$$

II. Let us assume that

$$
Y=x y
$$

This equation, taking (2) into account, takes the following form:

$$
Y=\frac{b x}{a} \sqrt{a^{2}-x^{2}}
$$

from which

$$
Y^{\prime}=\lim \frac{Y_{1}-Y_{0}}{x_{1}-x_{0}}=\frac{a^{2} b-2 b x^{2}}{a \sqrt{a^{2}-x^{2}}}
$$

The sign of the latter depends on the sign of

$$
\frac{a^{2}}{2}-x^{2}
$$

The following table shows the variations of $x y$:

$$
\frac{x \left\lvert\, o \frac{a \sqrt{2}}{2} a\right.}{Y^{\prime} \mid+o \quad-}
$$

At $x=x_{m}=\frac{a \sqrt{2}}{2}, x y=\frac{a b}{2}=\operatorname{maximum.~The~} x_{m}$ construction is very simple. From the midpoint $C$ of $O A$, as the center, we draw a circle with radius $O C=A C$. We raise a perpendicular to $O A$ at $C$. Any intersection point of this perpendicular and the circle is at the desired $x_{m}$ distance from $O$.