[Area and volume (extremum problems).]

In a regular quadrilateral pyramid, two identical spheres of radius $r$ are placed, touching the base of the pyramid at points lying on the segment connecting the midpoints of opposite sides of the base. Each sphere touches a lateral face of the pyramid and the other sphere. Find the height of the pyramid for which the volume of the pyramid is the smallest.
Let $PABCD$ be a regular quadrilateral pyramid with vertex $P$ (Fig.1); $M$ and $N$ are the midpoints of the sides $AD$ and $BC$ of the base $ABCD$, respectively. The section of the pyramid by the plane passing through points $P, M$, and $N$ is an isosceles triangle $PMN$, with the base $MN$ equal to the side of the square $ABCD$ (Fig.2).

Circles of radius $r$ touch the height $PQ$ at the same point $K$. The circle with center $O_1$ is inscribed in the angle $PMN$, and the circle with center $O_2$ is inscribed in the angle $PNM$. Let the side of the square $ABCD$ be $a$, the height of the pyramid be $x$, and the circle with center $O_1$ touch $MN$ at point $F$. Suppose $\angle PMN = \angle PNM = 2\alpha$. Then $\angle O_1MF = \alpha$. From the right triangles $PMQ$ and $O_1MF$, we find that

$$
\tan 2\alpha = \frac{PQ}{MQ} = \frac{2x}{a}, \quad \tan \alpha = \frac{Q_1F'}{MF'} = \frac{\frac{\varphi}{2} - \varphi}{2}
$$

Using the formula $\tan 2\alpha = \frac{2 \ln m^2}{1 - \lg^2 r_2}$, we get the equation

![](https://cdn.mathpix.com/cropped/2024_05_06_b09eb9152e3973547418g-51.jpg?height=143&width=243&top_left_y=1666&top_left_x=908)

from which we find that

$$
a = \frac{4r(x - \varphi)}{x - 2\varphi}
$$

If $V(x)$ is the volume of the pyramid $PABCD$, then

$$
V(x) = \frac{1}{3} S_{ABCD} \cdot PQ = \frac{1}{3} a^2 \cdot x = \frac{1}{3} \cdot \frac{16r^2 x (x - r)^2}{(x - 2r)^2}
$$

Since $x > 2r$, the problem reduces to finding the value of $x$ on the interval $(2r, +\infty)$ for which the function $V(x)$ attains its minimum value on this interval. Solving the equation $V'(x) = 0$, we find the critical points of the function $V(x)$. We consider only those that belong to the interval $(2r, +\infty)$.

$$
V'(x) = \frac{16r^2}{3} \cdot \frac{(x - \varphi)(x^2 - 5\psi x + 2r^2)}{x - 2\varphi}
$$

The interval $(2r, +\infty)$ contains the only root of this equation $x = \frac{r(5 + \sqrt{17})}{2}$. As we pass through the point $x = \frac{r(5 + \sqrt{17})}{2}$, the derivative changes sign from negative to positive. Therefore, on the interval $\left(2r, \frac{\varphi(5 + \sqrt{17})}{2}\right)$, the function $V(x)$ is decreasing, and on the interval $\left(\frac{r(5 + \sqrt{17})}{2}, +\infty\right)$, it is increasing. Thus, at $x = \frac{r(5 + \sqrt{17})}{2}$, the volume of the pyramid $PABCD$ is the smallest.

## Answer

The problem