Height of the Bottle

Relevant wiki: Volume of a Cylinder

We know that the volume of a right circular cylinder is area of the base multiplied by the height. Let $V_A$ be the volume of the figure on the left and $V_B$ be the volume of the figure on the right. We have

$V_A=\dfrac{\pi}{4}(16^2)(30-z)+\dfrac{\pi}{4}(4^2)(z)=\dfrac{\pi}{4}(256)(30-z)+\dfrac{\pi}{4}(16z)$

$V_B=\dfrac{\pi}{4}(4^2)(z+8)+\dfrac{\pi}{4}(16^2)(x-z-8)=\dfrac{\pi}{4}(16)(z+8)+\dfrac{\pi}{4}(256)(x-z-8)$

We know that $V_A=V_B$ , we can cancel out $\dfrac{\pi}{4}$ . Then equate $V_A$ and $V_B$ . So

$256(30-z)+16z=16(z+8)+256(x-z-8)$

$7680-256z+16z=16z+128+256x-256z-2048$

$7680=128+256x-2048$

$\boxed{x=37.5}$