A trapezoid has one base and two legs of unit length. Let's choose the length of the trapezoid's other base so that its area is as large as possible.
Solution. If the trapezoid is not symmetric, then it is a rhombus with unit sides, whose area is at most 1. In the case of a symmetric trapezoid, it can be assumed that the fourth side (the other base of the trapezoid) is greater than 1, since in the opposite case, the trapezoid can be enclosed in a larger area trapezoid by "flipping" the legs (see the figure).

![](https://cdn.mathpix.com/cropped/2024_05_02_4e002e4c3546e5749be4g-1.jpg?height=236&width=371&top_left_y=305&top_left_x=866)

Using the notation in the figure: the angle between the legs and the longer base is $\alpha$. Divide the trapezoid into a rhombus with unit sides and angles $\alpha$ and $\left(180^{\circ}-\alpha\right)$, and an isosceles triangle with unit legs and an angle of $\left(180^{\circ}-2 \alpha\right)$ between the legs; clearly, $0<\alpha \leq 90^{\circ}$; the case $\alpha=90^{\circ}$ corresponds to the unit square, which has the maximum area among the unit-sided rhombuses. The area of the trapezoid can be obtained as the sum of the areas of the isosceles triangle and the rhombus:

$$
T=\frac{1}{2} \sin \left(180^{\circ}-2 \alpha\right)+\sin \alpha=\frac{1}{2} \sin 2 \alpha+\sin \alpha=\sin \alpha(\cos \alpha+1)
$$

Express all of this in terms of $t=\operatorname{tg} \frac{\alpha}{2}$: $\sin \alpha=\frac{2 t}{1+t^{2}}, \cos \alpha=\frac{1-t^{2}}{1+t^{2}}$, thus

$$
T=\frac{2 t}{1+t^{2}} \cdot \frac{2}{1+t^{2}}=\frac{4 t}{\left(1+t^{2}\right)^{2}}
$$

According to the constraints on $\alpha$, $0<\frac{\alpha}{2} \leq 45^{\circ}$, so $0<t \leq 1$. By the arithmetic and geometric mean inequality,

$$
1+t^{2}=\frac{1}{3}+\frac{1}{3}+\frac{1}{3}+t^{2} \geq 4 \cdot \sqrt[4]{\frac{1}{3} \cdot \frac{1}{3} \cdot \frac{1}{3} \cdot t^{2}}=\frac{4 \cdot \sqrt{t}}{\sqrt[4]{27}}
$$

therefore

$$
T=\frac{4 t}{\left(1+t^{2}\right)^{2}} \leq \frac{4 t}{\left(\frac{4 \cdot \sqrt{t}}{\sqrt[4]{27}}\right)^{2}}=\frac{4 t}{\frac{16 t}{\sqrt{27}}}=\frac{3 \sqrt{3}}{4}
$$

This maximum value is taken by the area precisely when $\frac{1}{3}=t^{2}$, i.e., $\operatorname{tg} \frac{\alpha}{2}=t=\frac{1}{\sqrt{3}}=\operatorname{tg} 30^{\circ}$, or, since the tangent function is one-to-one on the interval $\left(0 ; 45^{\circ}\right]$, when $\alpha=60^{\circ}$.