7 Maximum distance on two spheres

[techy]

If you want a proof:
Assign the circle of radius 7 containing point P to have center R.
Assign the circle of radius 4 containing point Q to have center S.

P will always be 7 away from R, and Q will always be 4 away from S. However, their positions on the surface of the sphere may be changed.
     RP' = 7
     SQ' = 4
R and S are fixed points.
Therefore, the line segment PRQS must have length 7+7+4+4=22
     PRSQ= 22

Case 1:
If we move P to point P', this creates triangle P'QR. By triangle inequality, RQ+RP'>P'Q. However, RQ=7+8=15, and RP=7, so 22>P'Q. Therefore, moving P from the line segment PRQS will decrease the length of segment PQ in comparison to line segment PRQS.
     P'Q < 22

Case 2:
From symmetry, the same argument may be applied to the case where Q is moved to point Q'. 22>PQ'
     PQ' < PRSQ

Case 3:
In the case that both P and Q are moved to P' and Q' respectively, a quadrilateral emerges. In this case, we combine both previous cases. We have proved that P' will always move closer to Q, and Q' will always move closer to P. We use triangle inequality once again with triangle P'RQ':
RP'+RQ' > P'Q'
7+RQ' > P'Q'
We know that 7 <= RQ' <= 15 (7 when Q' is on the point of tangency and 15 when Q is on line segment RSQ).
Therefore, we take the maximum case of RQ'=15
7+15>P'Q'
22 > P'Q'
     P'Q' < PRSQ

Therefore, moving points P or Q from the line segment PRSQ will cause the segment PQ to be less than 22, which is the maximum
QED

[sriontube]

I thought the maximum distance will be along the circumference and it will be pi(7+4) which is more than 22.

[ceceliapassarella]

You also do not make me feel stupid or ashamed thanks

[markg]

That seemed too easy lol I always overthink questions like this when they're so easy and make them seem harder than they really are lol :/

[DouglasLenz]

6 people don't give two spheres about this shit