6. What is the smallest area that a right triangle can have, if its hypotenuse lies on the tangent to the graph of the function $y=\sqrt{x-3}$, one of its legs lies on the $y$-axis, and one of its vertices coincides with the point of tangency
Solution. $\quad f(x)=\sqrt{x-3}, \quad f^{\prime}\left(x_{0}\right)=\frac{1}{2 \sqrt{x-3}}$

$$
\begin{aligned}
& S_{A B C}=\frac{1}{2} A B \cdot B C, x_{0}-\text { abscissa of the point of tangency } A, \\
& A\left(x_{0}, f\left(x_{0}\right)\right), \quad B\left(0, f\left(x_{0}\right)\right), \quad C \quad \text { - intersection point }
\end{aligned}
$$

with the y-axis. Let $C(0, c)$. The equation of the tangent line to the graph of the function $f(x)=\sqrt{x-3}$ is $y=f^{\prime}\left(x_{0}\right)\left(x-x_{0}\right)+f\left(x_{0}\right)$. Point $C$ lies on the tangent line, so we substitute its coordinates into the equation of the tangent line: $c=-f^{\prime}\left(x_{0}\right) x_{0}+f\left(x_{0}\right)$.

Then $A B=x_{0}, \quad B C=f\left(x_{0}\right)-c=f^{\prime}\left(x_{0}\right) x_{0}, \quad S_{A B C}=\frac{1}{2} A B \cdot B C=\frac{1}{2} f^{\prime}\left(x_{0}\right)\left(x_{0}\right)^{2}=\frac{x_{0}^{2}}{2 \sqrt{x_{0}-3}}$. To find the extrema of the function $\quad S_{A B C}=S\left(x_{0}\right)=\frac{x_{0}^{2}}{4 \sqrt{x_{0}-3}}$ we find the zeros of the derivative of this function $S^{\prime}\left(x_{0}\right)=\frac{3\left(x_{0}^{2}-4 x_{0}\right)}{8 \sqrt{\left(x_{0}-3\right)^{3}}}$. Since $x_{0} \geq 3$, the only extremum point, and specifically, the point of minimum for this function is the point $\quad x_{0}=4, \quad S_{\min }=S(4)=\frac{4^{2}}{4 \sqrt{4-3}}=4$.

Answer: 4.