Problem 11. Find the maximum value of the function $y=3 x+4 \sqrt{1-x^{2}}$

$$
\text { S o l u t i o n. }
$$

First, let's find the domain of the given function: $D(f)=$ $=\left\{x \mid 1-x^{2} \geqslant 0\right\}=\{x \mid-1 \leqslant x \leqslant 1\} ; \quad E(f)=\{y \mid \quad$ the equation $y=3 x+4 \sqrt{1-x^{2}}$ has a solution on $\left.[-1 ; 1]\right\}=\{y \mid$ the equation $y-3 x=4 \sqrt{1-x^{2}}$ has a solution on $\left.[-1 ; 1]\right\}=\{y \mid$ the equation $25 x^{2}-6 x y+\left(y^{2}-16\right)=0, \quad$ where $y \geqslant 3 x$, has a solution on $[-1 ; 1]\}=$

$$
\begin{gathered}
=\left\{y \mid\left(y \geqslant 3 x \cap y^{2}-25 \leqslant 0\right) \cap-1 \leqslant x \leqslant 1\right\}= \\
=\{y \mid(y \geqslant 3 x \cap-5 \leqslant y \leqslant 5) \cap-1 \leqslant x \leqslant 1\}= \\
=\{y \mid-3 \leqslant y \leqslant 5\}
\end{gathered}
$$

16

Therefore, $\quad y_{\text {max }}=5$ (achieved when $\left.\quad x=\frac{3}{5}\right)$.

Let's explain the solution of the system $\left\{\begin{array}{l}y \geqslant 3 x \\ -5 \leqslant y\end{array}\right.$ $\left\{\begin{array}{l}-5 \leqslant y \leqslant 5 \\ -1 \leqslant x \leqslant 1\end{array}\right.$ graphically (Fig. 19).

3 p r o b l e m 12. A cylindrical rod is being turned. The length of the rod is $l$, the diameter of the base is $D, h$-the thickness of the layer being turned. Find the maximum displacement of the center of gravity during the turning of the cylinder ${ }^{1}$.

![](https://cdn.mathpix.com/cropped/2024_05_21_07b5a7f6f561b0ff634dg-18.jpg?height=705&width=439&top_left_y=270&top_left_x=1271)

Fig. 19

$$
\text { S o l u t i o n. }
$$

Introduce a rectangular coordinate system as shown in Fig. 20; $x$ - the length of the turned layer at some moment in time $t$ :

$$
0 \leqslant x \leqslant l
$$

Before turning, the center of gravity of the cylinder was at point $C\left(\frac{l}{2} ; 0\right)$. Then, as the turning progresses, point $C$ will start moving to the left. When a layer of thickness $h$ is turned from the entire surface of the cylinder, the center of gravity will return to its original position. At the beginning of the experiment, $x_{C}=\frac{l}{2}$, the mass of the cylinder $m=\frac{\pi D^{2} l}{4} \rho$, where $\rho$ is the density of the material the cylinder is made of.

After some time, the turning body will consist of two cylinders of different diameters: $D-2 h$ and $D$. The mass of the first cylinder $m_{1}=\frac{\pi(D-2 h)^{2} x}{4} \rho$, the abscissa of the center of gravity of the first cylinder $x_{1}=\frac{x}{2}$. The mass of the second cylinder $m_{2}=$ $=\frac{\pi D^{2}(l-x)}{4} \rho$, the abscissa of the center of gravity of the second cylinder $x_{2}=\frac{l-x}{2}+x=\frac{l+x}{2}$.[^0]

![](https://cdn.mathpix.com/cropped/2024_05_21_07b5a7f6f561b0ff634dg-18.jpg?height=434&width=460&top_left_y=1939&top_left_x=1255)

Fig. 20

Then the abscissa of the center of gravity of the system of two cylinders is

$$
\begin{gathered}
x_{C}=\frac{m_{1} x_{1}+m_{2} x_{2}}{m_{1}+m_{2}}= \\
=\frac{\frac{\pi(D-2 h)^{2} x^{2} \rho}{8}+\frac{\pi D^{2}(l-x) \rho(l+x)}{8}}{\frac{\pi(D-2 h)^{2} x \rho}{4}+\frac{\pi D^{2}(l-x) \rho}{4}}= \\
=\frac{(D-2 h)^{2} x^{2}+D^{2}\left(l^{2}-x^{2}\right)}{2\left((D-2 h)^{2} x+D^{2}(l-x)\right)}=\frac{4 h(h-D) x^{2}+D^{2} l^{2}}{8 h(h-D) x+2 D^{2} l}
\end{gathered}
$$

The problem has been reduced to finding the maximum value of the function

$$
y=\frac{4 h(h-D) x^{2}+D^{2} l^{2}}{8 h(h-D) x+2 D^{2} l}
$$

Let's find the range of values of this function:

$E(f)=\left\{y \left\lvert\, y=\frac{4 h(h-D) x^{2}+D^{2} l^{2}}{8 h(h-D) x+2 D^{2} l}\right.\right.$ has a solution in $[0 ; l[\}=$ $=\left\{y \mid 4(h-D) h x^{2}-8 h(h-D) y x+D^{2} l^{2}-2 D^{2} l y=0\right.$ has a solution in $\left[0 ; l[\}=\left\{y \left\lvert\,\left(y \geqslant \frac{D l}{2 h} \cup y \leqslant \frac{D l}{2(D-h)}\right) \cap 0 \leqslant y \leqslant l\right.\right\}=\{y \mid 0 \leqslant\right.$ $\left.\leqslant y \leqslant \frac{D l}{2(D-h)}\right\}$

Let's explain the last conclusion: $D>2 h \Rightarrow \frac{D}{2 h}>1 \Rightarrow \frac{D l}{2 h}>l$. Therefore, we have $0 \leqslant y \leqslant \frac{D l}{2(D-h)}$.

$$
\begin{gathered}
y_{\text {max }}=\frac{D l}{2(D-h)} ; \quad x_{\text {max }}=\frac{D l}{2(D-h)} \\
C C_{\text {max }}^{\prime}=\frac{D l}{2(D-h)}-\frac{l}{2}=\frac{l h}{2(D-h)}
\end{gathered}
$$

A series of problems reduces to finding the maximum or minimum value of a quadratic trinomial.

T h e o r e m. A quadratic trinomial $y=a x^{2}+b x+c$ has a maximum or minimum value, which is taken at $x=-\frac{b}{2 a}$; the value is minimum if $a>0$, and maximum if $a<0$. If $a>0$, the first term $a\left(x+\frac{b}{2 a}\right)^{2}$ cannot be negative; it becomes zero at $x=-\frac{b}{2 a}$. In this case, $y$ takes its minimum value $y=c-\frac{b^{2}}{4 a}$ and the function does not have a maximum value. If $a<0$, then $a\left(x+\frac{b}{2 a}\right)^{2} \leqslant 0$, but at $x=-\frac{b}{2 a}$ it becomes zero. Therefore, $y$ reaches its maximum value $y=c-\frac{b^{2}}{4 a}$. In this case, there is no minimum value.

Corollary. The product of two positive factors, whose sum is constant, reaches its maximum value when these factors are equal ${ }^{1}$.

Proof. Let $p$ be the sum of these two factors. If the first factor is $x$, then the second is $p-x$. The product of the considered factors $y=x(p-x)=-x^{2}+p x$, as follows from the proven theorem, takes its maximum value, equal to $y=\frac{p^{2}}{4}$, at $x=\frac{p}{2}$.