Lines: y=3x+10, y=2x+15, y=x+22. Slopes are decreasing (sorted). Build the lower envelope. Add line 1 (3x+10): deque = [L1]. Add line 2 (2x+15): check if L1 is still useful.
Intersection at x=5. L2 better for x>5. Deque = [L1, L2]. Add line 3 (x+22): L3 intersects L2 at x=7. L3 intersects L1 at x=6. L2 is dominated: remove it. Final deque = [L1, L3]. Query x=4: L1 gives 22, L3 gives 26. Min = 22 from L1. Deque structure gives O(1) amortized query.