The solution worth ~30\%~ of points is trivial and implements precisely what is required in the task: from all possible choices of divisions of potato bags in the stores, which is ~\mathcal O(2^N)~, choose the one with the minimal product of average prices where at least one half is of size ~L~. The complexity of this solution is ~\mathcal O(2^N \times N)~. Let's now try to solve the task for all points.

Let's denote the total price of the potatoes in the first store as ~c_1~ and the total number of potatoes as ~w_1~, and analogously in the second store as ~c_2~ and ~w_2~, the product of the average prices is equal to ~c_1 \times c_2 / w_1 / w_2~. Given the fact that the sum of prices and the sum of the number of potatoes is constant, we can modify this expression as ~c_1 \times (c-c_1) / (w_1 \times (w-w_1))~. If we fix the parameter ~w_1~, we can notice that this expression is minimal when ~c_1~ is minimal as well. Now our task is to find the minimal ~c_1~ for each ~w_1~ such that the first store contains exactly ~L~ or ~N-L~ bags of potatoes. The minimal such ~c_1~ is found using dynamic programming. Let ~f(n, w, l)~ be the smallest price when choosing exactly ~l~ bags of potatoes out of the first ~n~ bags of potatoes so that the chosen bags contain exactly ~w~ potatoes. The relation in this dynamic is left as an exercise to the reader. The total complexity of the solution is ~\mathcal O(wn^2)~.