12.45 In a right triangle with a hypotenuse of 24 cm and an angle of $60^{\circ}$, a rectangle is inscribed, with its base lying on the hypotenuse. What should be the lengths of the sides of the rectangle so that its area is maximized?
12.45 According to the condition, $AB=24 \text{ cm}, \angle A=60^{\circ}$, from which $\angle B=30^{\circ}$ (Fig. 12.7). Let $MN=x$ and $LM=y$ be the linear dimensions of the rectangle $MNKL$ (in centimeters). Express $x$ in terms of $y$. From $\triangle KBL: \quad BL=KL \operatorname{ctg} 30^{\circ}=x \sqrt{3}$; from $\triangle MNA$:
$AM=MN \operatorname{ctg} 60^{\circ}=x \frac{\sqrt{3}}{3}$; then

$y=ML=AB-BL-AM=24-x \sqrt{3}-x \frac{\sqrt{3}}{3}=24-\frac{4 x \sqrt{3}}{3}$.

The area of the rectangle $MNKL$ is expressed by the function $S(x)=x y=x\left(24-\frac{4 x \sqrt{3}}{3}\right)=24 x-\frac{4 x^{2} \sqrt{3}}{3}$. Further, we have $S^{\prime}(x)=24-\frac{8 \sqrt{3}}{3} x$, from which $S^{\prime}(x)=0$ when $x=3 \sqrt{3}$. If $x<3 \sqrt{3}$, then $S^{\prime}(x)>0$, and if $x>3 \sqrt{3}$, then $S^{\prime}(x)<0$, i.e., $x=3 \sqrt{3}$ is the point of maximum of the function $S(x)$. Thus, the lengths of the sides of the desired rectangle are $3 \sqrt{3}$ cm and $24-\frac{4 \sqrt{3}}{3} \cdot 3 \sqrt{3}=12 \text{ cm}$.

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-396.jpg?height=630&width=404&top_left_y=933&top_left_x=196)

Fig. 12.7

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-396.jpg?height=580&width=409&top_left_y=985&top_left_x=703)

Fig. 12.8

Answer: $3 \sqrt{3}$ and 12 cm.