1. A square with side $2a$ is cut into four isosceles triangles, whose bases are the sides of the given square and whose heights are equal to $x$. The remaining deck of the square represents the net of a regular quadrilateral pyramid.

a) Express the volume $V$ of this pyramid as a function of the parameter $x$.

b) Determine $x$ so that the volume $V$ is maximized.

c) Sketch the graph of the function $V(x)$.
Solution. a) We obtain a pyramid with a square base with a diagonal of $2 a-2 x$. It is easy to see that the lateral edge and the height of the pyramid are respectively equal to $\sqrt{a^{2}+x^{2}}$ and $\sqrt{2 a x}$. From this, for the volume of the pyramid, we get

$$
V(x)=\frac{1}{3} \frac{(2 a-2 x)^{2}}{2} \sqrt{2 a x}=\frac{2 \sqrt{2 a}}{3} \sqrt{x}(a-x)^{2}
$$

where $00$, if $0<x<\frac{a}{5}$,
- $V^{\prime}(x)=0$, if $x=\frac{a}{5}$,
- $V^{\prime}(x)<0$, if $\frac{a}{5}<x<a$.

According to this, the function $V(x)$ increases on the interval $\left(0, \frac{a}{5}\right)$ and decreases on the interval $\left(\frac{a}{5}, a\right)$, and has a maximum at the point $x=\frac{a}{5}$.

b) The second derivative of the function is

$$
V^{\prime \prime}(x)=\frac{2 \sqrt{2 a}}{3} \frac{15 x^{2}-6 a x-a^{2}}{4 x \sqrt{x}}, 0<x<a
$$

The quadratic trinomial $15 x^{2}-6 a x-a^{2}$ has a zero in the interval $(0, a)$ at $x_{0}=\frac{3+\sqrt{24}}{15} a$. This is an inflection point of the function $V(x)$ whose graph is shown in the figure on the right.

![](https://cdn.mathpix.com/cropped/2024_06_05_17a2a9aba963f6a4e511g-3.jpg?height=387&width=504&top_left_y=941&top_left_x=957)