( 429 ) 



So for the minimum value of (<p"') this equation must be equal to 0. 



. , d*A 



I his expression is equal to 0, if* = 0, or if: 



dx* 



n.r 1 — x 



,( dA\ ( dA 



The latter can only he the case, if in (he second member the 

 sign -f- is retained, and - is rejected. This means that in the 

 expression : 



( dA \ 



f dx \ 



1 — \/\A — x — 



only the sign -f must hold in the numerator of the second member, 

 b 3 



or that — > 1. So if the closed curve is restricted to volumes smaller 

 v 



than b 2 . 



If we seek the value of x which satisfies: 

 n\v 2 (l-#) 



dA dA 



A~x — A + (l-x) — 



ax dx 



then when this value of x is substituted, (<//") must be negative, 

 because (<p'") has proved to be positive for x = and x = l. 

 t For it is not sufficient for the existence of 2 roots of the equation 

 <p'" = 1 that if'" has a minimum value, but it is also required that 

 this minimum value is negative. 

 Tf we substitute in : 



A * dA \ A dA) 



n — \ — n |/ \A~ x — — |/ \A + (1— x) — 

 / dx \ dx \ 



the value of: 



then :. 



dA nx I dA 



1 — x \ dx | 



must be negative. 



Now we find from the condition on which y" has a minimum 



value : 



dA (l-«) s -nV 



= A 



dx x (1— #) {l—x-\-n?x\ 



