Optimal space coverage with white convex polygons

Assume that we are given a set of points some of which are black and the rest are white. The goal is to find a set of convex polygons with maximum total area that cover all white points and exclude all black points. We study the problem on three different settings (based on overlapping between different convex polygons): (1) In case convex polygons are permitted to have common area, we present a polynomial algorithm. (2) In case convex polygons are not allowed to have common area but are allowed to have common vertices, we prove the NP-hardness of the problem and propose an algorithm whose output is at least OPTlog(2n/OPT)+2log(n)1/4\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\left( \frac{OPT}{log(2n/OPT) + 2log(n)}\right) ^{1/4}$$\end{document}. (3) Finally, in case convex polygons are not allowed to have common area or common vertices, also we prove the NP-hardness of the problem and propose an algorithm whose output is at least 334.πOPTlog(2n/OPT)+2log(n)1/4\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\frac{3\sqrt{3}}{4.\pi }\left( \frac{OPT}{log(2n/OPT) + 2log(n)}\right) ^{1/4}$$\end{document}.