A pyramid has a rectangular base, its sides are equal to each other and to one of the base edges. By changing the other base edge, when will the volume of the pyramid be the largest?

 

The pyramid has a rectangular base, its sides are equal to each other and to one of the base edges. By changing the other base edge, when will the volume of the pyramid be the largest?
Let the slant height of the pyramid be $a$, the variable base edge $b$, and the height $m$.

![](https://cdn.mathpix.com/cropped/2024_05_02_e8be3e490d85be38994bg-1.jpg?height=475&width=681&top_left_y=218&top_left_x=711)

According to the Pythagorean theorem in our pyramid,

$$
m=\sqrt{a^{2}-\frac{a^{2}+b^{2}}{4}}=\frac{\sqrt{3 a^{2}-b^{2}}}{2}
$$

thus the volume is:

$$
V=\frac{a \cdot b \sqrt{3 a^{2}-b^{2}}}{6}
$$

This is only meaningful if $b^{2} \leq 3 a^{2}$.

By varying the value of $b$, the value of $V$ is maximal when the product $\left(\frac{6 V}{a}\right)^{2}=b^{2}\left(3 a^{2}-b^{2}\right)$ is maximal. This, due to the arithmetic and geometric mean inequality, is never greater than $\left(\frac{3 a^{2}}{2}\right)^{2}$, and equality holds when the two numbers are equal: $b^{2}=3 a^{2}-b^{2}, b=a \sqrt{\frac{3}{2}}=a \cdot 1.225$.

We still need to check whether a pyramid that meets the problem's conditions can indeed be formed with $b=\sqrt{\frac{3}{2}} a$. For this, the triangle inequality $2 a \geq b$ must be satisfied, which is indeed the case with the found value.