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

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Solution. $\quad f(x)=\sqrt{x-3}, \quad f^{\prime}\left(x_{0}\right)=\frac{1}{2 \sqrt{x-3}}$

$S_{A B C}=\frac{1}{2} A B \cdot B C, x_{0}-$ abscissa of the point of tangency $A$,

$A\left(x_{0}, f\left(x_{0}\right)\right), \quad B\left(0, f\left(x_{0}\right)\right), \quad C \quad$ - the point of intersection of the tangent with the $y$-axis. Let $C(0, c)$. The equation

![](https://cdn.mathpix.com/cropped/2024_05_06_58f34d7e66d046d9b7d5g-04.jpg?height=474&width=711&top_left_y=200&top_left_x=1278)
of the tangent 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 $C$ belongs to the tangent, we substitute its coordinates into the equation of the tangent: $c=-f^{\prime}\left(x_{0}\right) x_{0}+f\left(x_{0}\right)$.

Then $\quad A B=x_{0}, \quad B C=f\left(x_{0}\right)-c=f^{\prime}\left(x_{0}\right) x_{0}, \quad S_{A B C}=\frac{1}{2} A B \cdot B C=\frac{1}{2} f^{\prime}\left(x_{0}\right)\left(x_{0}\right)^{2}=\frac{x_{0}^{2}}{2 \sqrt{x_{0}-3}} . \quad$ To find the extrema of the function $S_{A B C}=S\left(x_{0}\right)=\frac{x_{0}{ }^{2}}{4 \sqrt{x_{0}-3}}$ we find the zeros of the derivative of this function $S^{\prime}\left(x_{0}\right)=\frac{3\left(x_{0}{ }^{2}-4 x_{0}\right)}{8 \sqrt{\left(x_{0}-3\right)^{3}}} . \quad$ 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.