Consider a polygon ABCDE with coordinates A(0,4), B(3,0), C(7,2) , D(11,6) and E(8,15)
Choose all possible pairs from below at which a maximum over the region can be attained by a non constant linear function.
[(0,4),(7,2)] [(0,4),(8,15)]
[(7,2),(11,6)] [(11,6),(8,15)]
[(0,4),(3,0)] [(3,0),(7,2)]
[(3,0),(11,6)] [(0,4),(11,6)]
[(7,2),(8,15)] [(3,0),(8,15)]
Can someone at least help get me started?
Thanks so much!
i don't no how to do it
16
To determine the pairs where a maximum over the region can be attained by a non-constant linear function, we need to find the pair of points that can form the longest segment.
To do this, we can calculate the distance between each pair of points using the distance formula: d = √((x2-x1)^2 + (y2-y1)^2).
Let's calculate the distances between each pair of points:
1. Distance between (0,4) and (7,2):
d1 = √((7-0)^2 + (2-4)^2) = √49 + 4 = √53
2. Distance between (0,4) and (8,15):
d2 = √((8-0)^2 + (15-4)^2) = √64 + 121 = √185
3. Distance between (7,2) and (11,6):
d3 = √((11-7)^2 + (6-2)^2) = √16 + 16 = √32
4. Distance between (11,6) and (8,15):
d4 = √((8-11)^2 + (15-6)^2) = √9 + 81 = √90
5. Distance between (0,4) and (3,0):
d5 = √((3-0)^2 + (0-4)^2) = √9 + 16 = √25
6. Distance between (3,0) and (7,2):
d6 = √((7-3)^2 + (2-0)^2) = √16 + 4 = √20
7. Distance between (3,0) and (11,6):
d7 = √((11-3)^2 + (6-0)^2) = √64 + 36 = √100
8. Distance between (0,4) and (11,6):
d8 = √((11-0)^2 + (6-4)^2) = √121 + 4 = √125
9. Distance between (7,2) and (8,15):
d9 = √((8-7)^2 + (15-2)^2) = √1 + 169 = √170
10. Distance between (3,0) and (8,15):
d10 = √((8-3)^2 + (15-0)^2) = √25 + 225 = √250
By comparing the distances, we can see that the pairs [(0,4),(3,0)] and [(3,0),(7,2)] have the longest segments, yielding the maximum over the region. Thus, the answer is [(0,4),(3,0)] and [(3,0),(7,2)].