136 Remarks on Several Subjecis. 
. ihe . tne 
= V¥1+tt whence T?=¢ +0, whose differential put 
=0 gives t=/9, whence c=yt=y/9. : 
Probiem 2. Givenu=1ty?x to be a maximum, when v= 
ay (o/yy+xrr+y) is constant. 
Putting cyt, we get u=1 4. 3, t, pany? (/1 +tt+-1) 
whence m=3, n=2, and, (neglecting the constant re 
Pimet: T’=4/1 +t Pee erin ea 
r 
faite ase (1+ + 1)3 
=f/t%+1% +f ? whose differential put=o and reduced, 
gives for the maximum of u, t=2,/2, whence c=y. 2/2. — 
3 
Problem 8. Suppose eat eae Ct (p? + 4px)? — 
~* t, _p being the parameter o a parabola, whose equa- 
quation is pr=yy (p bein ng Salted a by Professor Fisher. ) 
Then v being given itis required to find ua maximum. 
Putting r=pl, we getuxF. p?s t%) od, wept. J (1+ 
4t) Fy f: [p taking the place of the unknown quantity 
x ory in the forms treated of above.] Hence m=3, nai 
and neglecting the constant factors T/=¢? ; T’”=(1 +40)? 
2 
t iS 
—1, whence Ten, = 3 35 whenceT* 
a rv (1449? —1 ; i 
ss he 
— grew : 
=e 3(1-441) 2 3, whose differential put =o, and 
reduced gives 4t¢—15t+12=o0 whence tat oe a 
ae <s e 
er Spee 
The solution of this problem by any usual m 
seems to have been considered as quite tedious by Profes 
sor Fisher, in page 88 Vol. V. of the Journal. 
Problem 14. ‘Suppose u=2z sin, y to be a maximum 
when. v=x*.yis a given quantity. 
Here the quantities require no previous preparation. 
For by putting m=1. n=2, T’/=2sin y, T’=y, we e get 
