## Task B-2.4.

The perimeter of a right trapezoid is $20 \mathrm{~cm}$, and the cosine of its acute angle is $\frac{\sqrt{5}}{3}$. What should the height of the trapezoid be to maximize its area? What is the maximum area?
## Solution.

![](https://cdn.mathpix.com/cropped/2024_05_30_8cadc17e7912dc7108beg-08.jpg?height=437&width=705&top_left_y=1432&top_left_x=630)

$\sin \alpha = \sqrt{1 - \cos^2 \alpha} = \frac{2}{3}$. 1 point

From $\sin \alpha = \frac{v}{d}$, we get $d = \frac{v}{\sin \alpha} = \frac{3}{2} v$.

Then from the given perimeter, we have

$o = a + c + v + d \Rightarrow a + c = o - (v + d) = 20 - \frac{5}{2} v$. 1 point

$P = \frac{a + c}{2} \cdot v = \frac{-5}{4} v + 10 v$.

Since the area of the trapezoid is a quadratic function of the height, the maximum is achieved at the vertex $\left(x_{0} = \frac{-b}{2a}\right)$.

$v = \frac{-10}{\frac{-5}{2}} = 4$.

The trapezoid will have the maximum area for a height of $4 \text{ cm}$.

The maximum area is $20 \text{ cm}^2$.