beaker needs water

Let the center of gravity be $h$ mm from the bottom of the beaker and the depth of water be $x$ . Then $h$ is given by: $h=\frac{m_wh_w+m_bh_b}{m_w+m_b}$ where $m_w$ and $h_w$ , and $m_b$ and $h_b$ are mass and the height of the center of gravity from the beaker bottom of water and those of the beaker respectively.

It is noted that $m_w=\rho \pi (30^2)x$ , where $\rho = 1$ $g/cm^3$ or $1$ $g/1000$ $mm^3$ is the density of water. $\Rightarrow m_w = 0.9\pi x$ . Also that $h_b=\frac{x}{2}$ . Therefore, $h=\frac{0.45\pi x^2+10000}{0.9\pi x+100}$ For the beaker to be as stable as possible, $h$ is minimum, which occurs when $\frac{dh}{dx}=0$ . $\frac{dh}{dx}=\frac{0.9\pi x}{0.9\pi x+100}-\frac{0.9\pi (0.45\pi x^2+10000)}{(0.9\pi x+100)^2}$ $=\frac{0.9\pi (0.45\pi x^2+100x-10000)}{(0.9\pi x+100)^2}$ When $\frac{dh}{dx}=0$ , $\Rightarrow 0.45\pi x^2+100x-10000=0$ . Solving the quadratic equation, we get $x=55.87\approx \boxed {56}mm$ .