## Task B-3.7.

In a right-angled triangle, the sum of the lengths of the legs and three times the length of the height on the hypotenuse is 12. The triangle rotates around the line on which the hypotenuse lies, thus forming a rotational body of maximum surface area. What is the volume of the resulting body?
## Solution.

By rotating a right-angled triangle around its hypotenuse, a body is formed consisting of two cones with a common base whose radius is equal to the height to the hypotenuse.

![](https://cdn.mathpix.com/cropped/2024_05_30_bbee2d45ecbf3fd03f8eg-17.jpg?height=417&width=406&top_left_y=414&top_left_x=748)

The surface area consists of two lateral surfaces whose slant heights are the legs \(a\) and \(b\) of the given right-angled triangle. The radius of the base of the cone is the height \(v\) of the given right-angled triangle. Therefore, we have

\[
\begin{gathered}
a + b + 3v = 12 \\
O = v \pi a + v \pi b = v \pi (12 - 3v) = 3 \pi (-v^2 + 4v)
\end{gathered}
\]

The surface area will be maximal for the \(v\) at which the function \(f(v) = -v^2 + 4v\) achieves its maximum, which is for

\[
v = -\frac{4}{-2} = 2
\]

From \(a + b = 12 - 3v\) it follows that \(a + b = 6\) (1). 1 point

From \(\frac{ab}{2} = \frac{cv}{2}\) it follows that \(ab = cv = 2c\) (2). 1 point

By squaring equation (1) we get \((a + b)^2 = 36, a^2 + 2ab + b^2 = 36\).

Then, due to (2) and the Pythagorean theorem \((a^2 + b^2 = c^2)\), the last equation becomes \(c^2 + 4c - 36 = 0\). 2 points

1 point

The positive solution to the obtained quadratic equation is \(c = -2 + 2\sqrt{10}\).

Finally, we get \(V = \frac{8}{3} \pi (\sqrt{10} - 1)\) cubic units. 1 point

Note: For the first point, it is sufficient to draw a sketch or describe the body that is formed in words.

## COUNTY COMPETITION IN MATHEMATICS

4th grade - high school - B variant
February 5, 2016.

## IF A STUDENT HAS A DIFFERENT APPROACH TO SOLVING THE TASK, THE COMMITTEE IS OBLIGED TO GRADE AND EVALUATE THAT APPROACH APPROPRIATELY.