25-26高三上·江苏淮安涟水一中月考·第14题

$\frac{9(\sqrt{2}+\sqrt{6})}{2}$

如图，因$S_{\triangle ABC} = S_{\triangle ABD} + S_{\triangle BCD}$，
则可得$\frac{1}{2}ac\sin150^\circ = \frac{1}{2}(a + c)\sin75^\circ$，
即$\frac{1}{2}ac = (a + c)\sin(30^\circ + 45^\circ)$$= \frac{\sqrt{2} + \sqrt{6}}{4}(a + c)$，
化简得$\frac{1}{a} + \frac{1}{c} = \frac{\sqrt{2} + \sqrt{6}}{\sqrt{2}\sqrt{6}}$，
因$a, c > 0$，
则$4a + c = \frac{\sqrt{2} + \sqrt{6}}{2}(4a + c)\left(\frac{1}{a} + \frac{1}{c}\right)$$= \frac{\sqrt{2} + \sqrt{6}}{2}\left(5 + \frac{4a}{c} + \frac{c}{a}\right)$$\geq \frac{\sqrt{2} + \sqrt{6}}{2}\left(5 + 2\sqrt{4}\right) = \frac{9(\sqrt{2}+\sqrt{6})}{2}$，
当且仅当$\frac{4a}{c} = \frac{c}{a}$时，即$a = \frac{3}{4}(\sqrt{2}+\sqrt{6})$，$c = \frac{3}{2}(\sqrt{2}+\sqrt{6})$时，取等号，
故$4a + c$的最小值为$\frac{9(\sqrt{2}+\sqrt{6})}{2}$．

故答案为：$\frac{9(\sqrt{2}+\sqrt{6})}{2}$.