Determine the maximum of the function

$$
f(t)=a \sin t+b \cos t
$$

in the interval $0<t<2 \pi$. ( $a$ and $b$ are constants.)
I. solution: We can transform $f(t)$ into a constant multiple of the sine of a single angle if we find a constant $\lambda$ such that in the transformation

$$
\mathrm{f}(t)=\lambda\left(\frac{a}{\lambda} \sin t+\frac{b}{\lambda} \cos t\right)
$$

$\frac{a}{\lambda}$ and $\frac{b}{\lambda}$ are the sine and cosine of the same angle; for this, it is only necessary that the sum of their squares be 1, that is,

$$
\lambda=\sqrt{a^{2}+b^{2}}
$$

In this case (taking into account the sign of $a$ and $b$ as well) there is a uniquely determined angle $\varphi$ between 0 and $2 \pi$ such that

$$
\cos \varphi=\frac{a}{\sqrt{a^{2}+b^{2}}}, \quad \sin \varphi=\frac{b}{\sqrt{a^{2}+b^{2}}}
$$

and

$$
\mathrm{f}(t)=\sqrt{a^{2}+b^{2}} \sin (t+\varphi)
$$

This takes its maximum when

$$
\sin (t+\varphi)=1
$$

that is, for $t$ between 0 and $2 \pi$, when $\varphi \leq \frac{\pi}{2}$, if

$$
t=\frac{\pi}{2}-\varphi
$$

$\frac{\pi}{2}>0$. The parametric equation of the centrally located ellipse: $x=b \cos t, y=a \sin t$, if the length of the axis on the $X$ axis is $2 b$, and the length of the axis on the $Y$ axis is $2 a$. Our task is to determine the maximum of the sum of the coordinates of the points of the ellipse.

The geometric locus of points whose coordinates sum to a constant $d$ is a line whose equation is $y=-x+d$. We need to find the largest $d$ value for which the above line still has a common point with the ellipse $\frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}}=1$, that is, we need to determine the $d$ value for which the line $y=-x+d$ is tangent to the ellipse $\frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}}=1$. Substitute $y-t$ into the equation of the ellipse and arrange it according to the powers of $x$:

$$
\left(a^{2}+b^{2}\right) x^{2}-2 b^{2} d x+b^{2} d^{2}+a^{2} b^{2}=0
$$

The line in question is a tangent if the quadratic equation has two equal roots, that is, if its discriminant is zero, that is,

$$
4 b^{4} d^{2}+4 a^{4} b^{2}-4 a^{2} b^{2} d^{2}+4 a^{2} b^{4}-4 b^{4} d^{2}=0
$$

From this, since $4 a^{2} b^{2} \neq 0, \quad a^{2}+b^{2}=d^{2}$.

Based on this,

$$
\mathrm{f}(t)_{\max }=(a \sin t+b \cos t)_{\max }=(y+x)_{\max }=(d)_{\max }=\sqrt{a^{2}+b^{2}}
$$

Wollner Róbert (Szeged, Radnóti g. IV. o. t.)

III. solution: The location of the extreme value of the function does not change if we square the function:

$$
\begin{aligned}
y & =\mathrm{f}^{2}(t)=a^{2} \sin ^{2} t+2 a b \sin t \cos t+b^{2} \cos ^{2} t= \\
& =a^{2}\left(1-\cos ^{2} t\right)+2 a b \sin t \cos t+b^{2}\left(1-\sin ^{2} t\right)= \\
& =a^{2}+b^{2}-(a \cos t-b \sin t)^{2}
\end{aligned}
$$

The maximum of the function $y(t)$ is where the expression in parentheses is zero, and the maximum is then $y(t)_{\max }=a^{2}+b^{2}$. Therefore, the maximum of the function we are looking for is

$$
\mathrm{f}(t)_{\max }=\sqrt{a^{2}+b^{2}}
$$

Szatmári Zoltán (Bp. VIII., Piarista g. IV. o. t.)