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University:  University of California, Merced - UCM

Greek Organization:  Kappa Sigma

Author:  ASIP

Comment:  For the � x terms: � 2 − 2 � x 2 −2x We add and subtract ( 2 2 ) 2 = 1 ( 2 2 ​ ) 2 =1 to complete the square. This gives us � 2 − 2 � + 1 − 1 = ( � − 1 ) 2 − 1 x 2 −2x+1−1=(x−1) 2 −1. For the � y terms: � 2 − 6 � y 2 −6y We add and subtract ( 6 2 ) 2 = 9 ( 2 6 ​ ) 2 =9 to complete the square. This gives us � 2 − 6 � + 9 − 9 = ( � − 3 ) 2 − 9 y 2 −6y+9−9=(y−3) 2 −9. Rewriting the Function Substituting these into the original function, we get: � ( � , � ) = ( � − 1 ) 2 − 1 + ( � − 3 ) 2 − 9 f(x,y)=(x−1) 2 −1+(y−3) 2 −9 This can be further simplified to: � ( � , � ) = ( � − 1 ) 2 + ( � − 3 ) 2 − 10 f(x,y)=(x−1) 2 +(y−3) 2 −10 Analyzing the Function The squared terms ( � − 1 ) 2 (x−1) 2 and ( � − 3 ) 2 (y−3) 2 are always non-negative since any real number squared is non-negative. The smallest value these squares can take is 0, which occurs when � = 1 x=1 and � = 3 y=3. Thus, the smallest value of � ( � , � ) f(x,y) is − 10 −10, achieved at the point ( 1 , 3 ) (1,3). There is no upper bound for � ( � , � ) f(x,y) since the squared terms can grow indefinitely as � x and � y move away from 1 and 3, respectively. Conclusion The smallest value of � ( � , � ) f(x,y) is − 10 −10, occurring at ( 1 , 3 ) (1,3). The function � ( � , � ) f(x,y) has no largest value; it grows indefinitely as � x and � y move away from 1 and 3, respectively. User can you just show the math no need for an explnation such as if it was for hw ChatGPT Certainly! Here's the concise mathematical work for the problem: Problem: Find the largest and smallest values of � ( � , � ) = � 2 − 2 � + � 2 − 6 � f(x,y)=x 2 −2x+y 2 −6y. Solution: Complete the square for � x: � 2 − 2 � = ( � − 1 ) 2 − 1 x 2 −2x=(x−1) 2 −1 Complete the square for � y: � 2 − 6 � = ( � − 3 ) 2 − 9 y 2 −6y=(y−3) 2 −9 Rewrite � ( � , � ) f(x,y): � ( � , � ) = ( � − 1 ) 2 − 1 + ( � − 3 ) 2 − 9 f(x,y)=(x−1) 2 −1+(y−3) 2 −9 Simplify: � ( � , � ) = ( � − 1 ) 2 + ( � − 3 ) 2 − 10 f(x,y)=(x−1) 2 +(y−3) 2 −10 Analyze the function: Minimum: � ( � , � ) f(x,y) is minimized when ( � − 1 ) 2 = 0 (x−1) 2 =0 and ( � − 3 ) 2 = 0 (y−3) 2 =0, i.e., at ( � , � ) = ( 1 , 3 ) (x,y)=(1,3). Minimum value: � ( 1 , 3 ) = 0 + 0 − 10 = − 10 f(1,3)=0+0−10=−10. No maximum value since ( � − 1 ) 2 (x−1) 2 and ( � − 3 ) 2 (y−3) 2 can grow indefinitely. Results: Smallest value of � ( � , � ) f(x,y) is − 10 −10, at ( 1 , 3 ) (1,3). No largest value for � ( � , � ) f(x,y).\n\n\n\n\n