W31] An Interesting Maximal Case 69 



Substituting the parametric values of x and y into the equation 



U = 4xy, we have 



U = 4a cos (|). bs in 4) = 2ab sin 2(|) . . . . (1) 



To find the critical value of ^ for a maximum value of U, set the 

 first derivative of U with respect to ^ equal to zero: 



whence 



Accordingly 



dU 



-T- = 4ab cos 2© = 0; 



dq) 



2^ = 90° or 270° 

 (|) = 45° or 135° 



a 



X = + —7=^ ; y = + 



Now 



—7- = — 8 ab sin 2 cb, 

 dcp- 



giving a negative value for cj) = 45°, but a positive value for ^ = 

 135°. Hence ^ = 45° is the critical value for a maximum. Hence 

 the only admissible values for giving a maximum are 



X = + 



y = + 



b 



V2' 



Substituting these values in (1), we have 



U = 4xy = 2ab, 



from which 



ab 

 xy = — . . . (2) 



The form in which this result presents itself, namely that of a 

 rectangular hyperbole xy = K, excites interest at once, and prompts 

 a further investigation. The introduction into the problem of an 

 apparently extraneous curve, not invoked in the original statement, 

 suggests that this locus must play a crucial role in connection with 

 the ellipse in the solution of the problem. 



a+b — a— b 



The co-ordinates + .— - , ,-— and ,— - , ■ ,— of two of the 



v 2 v 2 V 2 v 2 



