Hmmm. Assuming a cylindrical bottle-interior, let it be centered on the z-axis with its bottom at z=0. WLOG we can normalize the (empty) bottle weight to 1, and assume its center of gravity to be at z=h. Then there is a constant C such that the weight of the liquid, filled to a height of z=t, is Ct in normalized units. Thus the center of gravity of the bottle filled to height t is z(t) = (1*h + Ct*(t/2)) / (1 + Ct) z'(t) = ((1+Ct)*(Ct) - (h + (C/2)t^2)(C)) / (1 + Ct)^2, which is 0 when Ct^2 + 4t - 2h = 0, or t = (-1 + sqrt(4 + 2hC)) / C. So the minimum center of gravity is at h + (-1 + sqrt(4 + 2C*h))^2 / 2C z = -------------------------------- sqrt(4 + 2C*h) Unless, as usual, I made a mistake. --Dan rwg wrote: << . . . a math-fun tie-in is to find, given the bottle weight and density of the "fluid", the fill depth which minimizes the height of the center of gravity.

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