Keep my drink stable

General solution

The glass has mass $\gamma m$ and its center of mass lies at height $\alpha L$ .

The drink has mass $x m$ with its center of mass in the middle, at height $\tfrac12x L$ .

The weighted average gives the height of the center of mass: $\frac{h(x)}L = \frac{\gamma m\cdot \alpha + x m \cdot \tfrac12 x}{\gamma m + x m} = \frac{\gamma\alpha + \tfrac12x^2}{\gamma + x}.$ This is minimal if its derivate is zero. Thus $\begin{aligned} 0 & = \frac{d}{dx} \frac{h(x)}L \\ 0 & = \frac{x(\gamma + x) - (\gamma\alpha + \tfrac12x^2)}{(\gamma + x)^2} \\ 0 & = x(\gamma + x) - (\gamma\alpha + \tfrac12x^2) \\ 0 & = \tfrac12 x^2 + \gamma x - \gamma\alpha \\ 0 & = x^2 + 2\gamma x - 2\gamma \alpha \\ \gamma^2 + 2\gamma\alpha & = (x + \gamma)^2 \\ x & = \sqrt{\gamma(\gamma + 2\alpha)} - \gamma. \end{aligned}$

With the given values, $x = \sqrt{0.25\cdot (0.25 + 2\cdot 0.40)} - 0.25 = \sqrt{0.2625} - 0.25 \approx \boxed{0.262}.$ Thus, the glass is filled for 26.2%.