Problem 15. A segment of a given length moves so that its ends slide along the sides of a right angle. In what position of this segment will the area of the cut-off triangle be the largest?

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\text { Solution. }
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Let the ends of the segment $A B$ of length $a$ slide along the sides of the angle $A C B: \widehat{A C B}=90^{\circ}$ (Fig. 21). Let $|A C|=x, 0 \leqslant x \leqslant a$. Then $|C B|=\sqrt{a^{2}-x^{2}}$. The area of triangle $A B C$ is $S=$ $=\frac{1}{2}|A C| \cdot|C B|=\frac{1}{2} x \sqrt{a^{2}-x^{2}}=\frac{1}{2} \sqrt{x^{2}\left(a^{2}-x^{2}\right)}$.

The sum of the non-negative terms $x^{2}$ and $a^{2}-x^{2}$ is constant (equal to $a^{2}$). Therefore, the product $x^{2}\left(a^{2}-x^{2}\right)$, and thus the area of the triangle $S$, reaches its maximum value when $\boldsymbol{x}^{\mathbf{2}}=a^{2}-x^{2}, \quad$ from which $\quad x=\frac{a \sqrt{2}}{2},|C B|=\frac{a \sqrt{2}}{2}$. Consequently, $\triangle A B C$ is isosceles:

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\widehat{A B C}=\widehat{C A B}=45^{\circ}, S=\frac{1}{4} a^{2}
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