= Solution
Let $m$ be the ice mass for which the stated drag coefficients are defined, $K=k_1+k_2$ and $f=2\Omega\sin\ell$. With stationary water, the steady horizontal ice equations are
$$
0=mf v_y+k_1v_{a,x}-Kv_x,\qquad0=-mf v_x+k_1v_{a,y}-Kv_y.
$$
The <linear system> gives
$$
v_y=\frac{k_1}{K^2+(mf)^2}\left(Kv_{a,y}-mf v_{a,x}\right).
$$
For fixed wind speed $V=10\,\mathrm{m\,s^{-1}}$, the <Cauchy-Schwarz inequality> minimizes this component when
$$
\boxed{\mathbf v_a=\frac{V}{\sqrt{K^2+(mf)^2}}(mf,-K),\qquad v_{y,\min}=-\frac{k_1V}{\sqrt{K^2+(mf)^2}}.}
$$
Thus the wind must blow towards a direction east of south by $\tan^{-1}(mf/K)$ in the northern hemisphere, west of south in the southern hemisphere, and due south at the equator. This compensates the <Coriolis force>; at the optimum $v_x=0$ and the ice moves due south. If the drag coefficients are per unit mass, replace $K/m$ by the corresponding total drag rate. “Blows towards” avoids the opposite convention used when naming winds by their origin.
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