2. Among all isosceles trapezoids where the angle at the base is $60^{\circ}$ and the area is equal to $6 \sqrt{3}$, determine the one that has the minimum perimeter.
2. First solution. Let $a$ and $b$ be the larger and smaller base of the trapezoid, $H$ its height, $\alpha$ the angle at the base, $P$ the area, and $O$ the perimeter. We have $a=b+2 H \operatorname{ctg} \alpha$, and from this $P=H \frac{a+b}{2}=H(b+H \operatorname{ctg} \alpha)$. Further, since the legs of this trapezoid are equal to $\frac{H}{\sin \alpha}$, we have $O=\frac{2 H}{\sin \alpha}+2 b+2 H \operatorname{ctg} \alpha=$ $2\left(\frac{H}{\sin \alpha}+b+H \operatorname{ctg} \alpha\right)$. From the previous formula for the area, we have $b+H \operatorname{ctg} \alpha=$ $\frac{P}{H}=\frac{6 \sqrt{3}}{H}$, so it follows that $O=2\left(\frac{H}{\sin \alpha}+\frac{6 \sqrt{3}}{H}\right)=$ $2\left(\frac{2 H}{\sqrt{3}}+\frac{6 \sqrt{3}}{H}\right)$, and the problem reduces to

![](https://cdn.mathpix.com/cropped/2024_06_04_f254ba4f25ce7f8d9df1g-46.jpg?height=384&width=589&top_left_y=1500&top_left_x=941)

Finding the minimum of the expression in parentheses, as required by 2017 4B 2.

If we denote this expression by $f(H)$, the derivative with respect to $H$ is $f^{\prime}(H)=\frac{2}{\sqrt{3}}-\frac{6 \sqrt{3}}{H^{2}}$. Setting this to zero, we get $\frac{2}{\sqrt{3}}=\frac{6 \sqrt{3}}{H^{2}}$, so $H^{2}=9$, and thus $H=3$ (since $H>0$). At this point, the minimum is indeed achieved because we see that $f^{\prime}(H) < 0$ for $H < 3$ and $f^{\prime}(H) > 0$ for $H > 3$. Then we get $b=\frac{6 \sqrt{3}}{3}-3 \operatorname{ctg} 60^{\circ}=2 \sqrt{3}-3 \frac{\sqrt{3}}{3}=\sqrt{3}, a=b+2 H \operatorname{ctg} \alpha=3 \sqrt{3}$, and $O=8 \sqrt{3}$.

Second solution. Using the same notation as in the first solution, let $c$ be the leg. Since $\alpha=60^{\circ}$, by observing the equilateral triangle above one leg, we get $a=b+c$ and $H=\frac{c \sqrt{3}}{2}$. Therefore, we have

$$
6 \sqrt{3}=P=\frac{c \sqrt{3}}{2} \cdot \frac{a+b}{2}=\frac{\sqrt{3}}{4} c(2 b+c)
$$

so we conclude $c(2 b+c)=24$. Further, we have $O=a+b+2 c=2 b+3 c$, and applying the inequality between the arithmetic and geometric means, we get

$$
\frac{2 b+3 c}{2}=\frac{(2 b+c)+(2 c)}{2} \geqslant \sqrt{(2 b+c)(2 c)}=\sqrt{48}
$$

i.e., $O \geqslant 2 \sqrt{48}=8 \sqrt{3}$. Equality is indeed achieved for $2 b+c=2 c$, i.e., $c=2 b$, so the minimum value of the perimeter is $8 \sqrt{3}$.