Worst-Case Optimal Covering of Rectangles by Disks

We provide the solution for a fundamental problem of geometric optimization by giving a complete characterization of worst-case optimal disk coverings of rectangles: For any $\lambda\geq 1$, the critical covering area $A^*(\lambda)$ is the minimum value for which any set of disks with total area at least $A^*(\lambda)$ can cover a rectangle of dimensions $\lambda\times 1$. We show that there is a threshold value $\lambda_2 = \sqrt{\sqrt{7}/2 - 1/4} \approx 1.035797\ldots$, such that for $\lambda<\lambda_2$ the critical covering area $A^*(\lambda)$ is $A^*(\lambda)=3\pi\left(\frac{\lambda^2}{16} +\frac{5}{32} + \frac{9}{256\lambda^2}\right)$, and for $\lambda\geq \lambda_2$, the critical area is $A^*(\lambda)=\pi(\lambda^2+2)/4$; these values are tight. For the special case $\lambda=1$, i.e., for covering a unit square, the critical covering area is $\frac{195\pi}{256}\approx 2.39301\ldots$. The proof uses a careful combination of manual and automatic analysis, demonstrating the power of the employed interval arithmetic technique.