2. What is the smallest area that a figure on the plane $Oxy$ can have, located between the lines $x=-26$ and $x=2$, bounded below by the $x$-axis, and above by the tangent to the graph of the function $y=5+\sqrt{4-x}$ at the point of tangency with abscissa $x_{0}$, where $x_{0}$ lies in the interval $-26 \leq x_{0} \leq 2$?

(12 points)
Solution. Solution. We will write the equation of the tangent line to the graph of the function $y=5+\sqrt{4-x}$ at the point with abscissa $x_{0}$

$$
y_{\text {tan }}=-\frac{1}{2 \sqrt{4-x_{0}}}\left(x-x_{0}\right)+5+\sqrt{4-x_{0}}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_03a96cebf8326f3655efg-3.jpg?height=339&width=774&top_left_y=1835&top_left_x=1212)

The figure is a trapezoid with bases equal to $y_{\text {tan }}(-26)$ and $y_{\text {tan }}(2)$. The height of the trapezoid is 28. We have

$y_{\text {tan }}(-26)=\frac{1}{2 \sqrt{4-x_{0}}}\left(26+x_{0}\right)+5+\sqrt{4-x_{0}}, \quad y_{\text {tan }}(2)=\frac{1}{2 \sqrt{4-x_{0}}}\left(x_{0}-2\right)+5+\sqrt{4-x_{0}}$,
$S=14\left(y_{\text {tan }}(-26)+y_{\text {tan }}(2)\right)=14\left(\frac{x_{0}+12}{\sqrt{4-x_{0}}}+10+2 \sqrt{4-x_{0}}\right)=14\left(\frac{20-x_{0}}{\sqrt{4-x_{0}}}+10\right)$. We find the derivative $S^{\prime}=14\left(\frac{20-x_{0}}{\sqrt{4-x_{0}}}\right)^{\prime}=\frac{-7\left(x_{0}+12\right)}{\sqrt{\left(4-x_{0}\right)^{3}}}$. The function $S\left(x_{0}\right)$ takes its minimum value at $x_{0}=-12$, $S_{\text {min }}=S(-12)=504$.

Answer: 504.