50 Prof. Challis on an Extension of the 



equated to zero, gives 



>ffiyifyi, _,*P oyofyo =Q . .. - ., 



and it is evident that this equation must apply to each circle 

 separately. On applying it to the first circle, it will be seen that 

 the second term vanishes, because, at the point P, y 8y =0, and 



^° = 1, p being infinitely great; and that for the other 



extremity of the are, since y x does not vanish and 8y 1 is indeter- 

 minate, we must have p^O. The same equation represent- 

 ing the integral from one extremity to the other of the second 

 circular arc, the first term vanishes, because y l Sy 1 = 0, and 



. l - =lj and the other because p =0, y not vanishing, 



yi+/?i 2 



and Sy being indeterminate. We may hence infer, making the 

 arbitrary constant c equal to X, that the straight line and the 

 arcs are continuous at the points of junction, and that each arc 

 is a quadrant of a circle. According to this investigation the 

 maximum solid has the form stated in my paper in the March 

 Number, viz. a cylinder with hemispherical ends of the same 

 radius. But it is not true, as there said, that a solid of this 

 form is larger than any other solid of revolution having the 

 same amount of surface and the same length of axis. I have 

 given reasons above for concluding that the equation Ap — 

 gives a maximum, subject to the condition that the generating 

 line of the surface, so far as regards the curved part, is symme- 

 trical with respect both to the axis of revolution and to a per- 

 pendicular axis ; and this condition the line we have found fulfils. 

 The line which gives an absolute maximum can only be deter- 

 mined by means of the equation A=0. 



The foregoing reasoning sufficiently meets the objection ad- 

 duced by Mr. Todhunter at the end of his communication in the 

 June Number, namely, that the solution does not satisfy the 

 equation A = 0. The answer is that it does not profess to satisfy 

 this equation, but a different equation, Kp — 0, for which reason 

 it was necessary to take account of the factor p as well as the 

 factor A. This having been doue, the solution satisfies A=0 as 

 far as regards the circular arcs, and p = as far as regards the 

 straight line, and therefore satisfies the equation Ap = through 

 the whole extent. But I grant that as it does not satisfy A = 

 through the whole extent, it does not give an absolute maxi- 

 mum. To obtain such a maximum it is necessary and sufficient 

 that the integral be such as to satisfy A = throughout, because 



