136 Remarks on Several Subjects. 
T= = whence Te age as whose differential put 
=0 gives t=, whence c=yt=yV/ 9. 
Probiem 2. Givenu=19y?2z to be a maximum, when v= 
ay (/yy+cr+y) is constant. 
Putting c=yt, we get u=1 @. y3. t, very? (1 iy 
whence m=3, n=2, and, ce the Constant factors) 
Tat; Ms Viti + ik 2 ie or 4 ha 
aes (J 1-itt “KE 1)3 
= ft s+ 4+" 3 whose differeiitial put=o and reduced, 
gives for the maximum of u, f=2,/2, whence a=y. 24/2. 
Problem 8. Suppose u=i7px?, v=} (ps ape 
ae p being the parameter of a parabola, whose equa- 
quation is px=yy (p being called a by Professor Fisher.) 
Then v being given itis required to find wa maximum. 
Putting +=pl, we get u=F. p?. t?, v=). «. p?. (i+ 
bag 
4t)?—] f , [p taking the place of the unknown quantity 
zory in the forms treated ofabove.] Hence m=3, n=2, 
3 
and neglecting the coustant factors T’/=¢? ; T’=(1+4t)? 
ee) eae ria aad 
—1, whence T= ——_= 3 3, whenceT 3 
PS” S(¢ay?-1}? 
pa 3 
3, whose differential put =o, and 
aaa ms 
sata gel --41). 2 — 
: S 
reduced gives 4t¢—15¢4+-12=0 whence easy: and 
Ek aan 
bev 
The ee of this problem by any usual method 
seems to have been considered as quite tedious by Profes- 
sor Fisher, in page 88 Vol. V. of the Journal. 
Problem 14. Suppose u=2a sin. y to be a maximum 
when v=2?. yis a given quantity. 
Here the quantities require no previous preparation. 
For by putting m=1. n=2, T’=2sin y, T’=y, we get 
