1921] 



An Interesting Maximal Case 



63 



(7I-0|(;;t-0' 



— 2h2 



-(;7T + ^)<v/T 



h^ 



V 2 



— h 



+ 



+ h- + 



2 2 



s/2 



v/2 



h'" — 



= h2 



1 + 



x/2 

 \/ 2 



This is likewise a positive quantity, by similar reasoning to that given 



a . . . 



above. Hence x = + —7= is a critical value, which gives rise to a 



maximum. From the equation of the ellipse, the corresponding 



b . . . ... 



value of y is + y— Rejecting the negative value as giving rise to 



a negative area, we have for the value of the maximum area 

 a b 



V" 2 • •/ 2 



= 2ab. 



Method II 



Representing a critical value by A and Ax by h, the customary 

 conditions for a maximum are 



f (A — h) = positive quantity; 

 f (A + h) = negative quantity. 



As before, let 



f(x) = (aV — x'')M 



2a2x — 4x3 



f (x) = 



a^ — 2x= 



>(a2x2 — x^)3^ \/a2— x2 



Since A = + -^ we have 

 s/ 2 



