square pyramId inside a sphere

Brief outline of my work: For simplicity, I make the radius of the sphere equal to 1. I will scale up the volume of the pyramid by $a^3$ at the end. Let $h$ be the height of the pyramid, and let $d$ be half of the diagonal of the base square of the pyramid; then the constraint is $1=d^2+(h-1)^2$ . (Draw a cross section through the tip of the pyramid and the diagonal of the base.)The volume of the pyramid will be $V=\frac{2}{3}d^2\times{h}=\frac{2}{3}(2h^2-h^3)$ for $0\leq{h}\leq{2}$ .Taking the derivative and finding the critical points shows that the maximum is attained at $h=\frac{4}{3}$ , where $d^2=\frac{8}{9}$ . (It's the maximum since $V$ is 0 at the end points.)Thus $V=\boxed{\frac{64}{81}a^3}$

The sphere may be thought of as centered at the origin $0, 0, 0$ . Place the top of the pyramid at $0, 0, a$ and the base at height $z$ . The coordinates of the vertices of the base are $\pm x, 0, z$ and $0, \pm x, z$ , with $x^2 + z^2 = a^2$ .

The height of the pyramid is $a - z$ and the side of its base is $\sqrt{2}\cdot x$ , so that $V = \frac{1}{3}\cdot (a-z)\cdot \left(\sqrt{2}x\right) = \frac{2}{3}(1-z)x^2 = \frac{2}{3}(a-z)(a^2-z^2).$

To optimize, take the derivative and equate it to zero: $0 = \frac{dV}{dz} = \frac{2}{3}\left(-(a^2-z^2)-2z(a-z)\right) = -\frac{2}{3}(a-z)(a+z+2z).$ The solution $z = a$ makes no sense, so we set the last factor equal to zero: $0 = a+z+2z = a+3z \Longrightarrow z = -\frac{a}{3}.$ Therefore $V = \frac{1}{3}\left(a+\frac{a}{3}\right)\left(a^2-\frac{a^2}{3^2}\right) = \frac{1}{3}\cdot\frac{4a}{3}\cdot\frac{8a}{9} = \frac{64}{81}a.$