Bounds or no bounds?
Finf the maximum value of ( 1 − x ) ( 2 − y ) ( 3 − z ) ( x + 2 y + 3 z ) , where x < 1 , y < 2 , z < 3 , and x + 2 y + 3 z > 0 . If the maximum value can be expressed as 2 b 3 a , find a b .
The answer is 35.
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1 solution
Do you not need to show that there exists ( x , y , z ) for which f ( x ) = 2 7 3 5 ? Otherwise we've found an upper bound, but not necessarily a maximum.
Yes, I forgot to add that, equality occurs when the values being averaged are equal. (I'll edit now)
f ( x , y , z ) = ( 1 − x ) ( 2 − y ) ( 3 − z ) ( x + 2 y + 3 z ) = 6 [ ( 1 − x ) ( 1 − 2 y ) ( 1 − 3 z ) ( x + 2 y + 3 z ) ] .
Using AM-GM inequality: [ ( 1 − x ) ( 1 − 2 y ) ( 1 − 3 z ) ( x + 2 y + 3 z ) ] 4 1 ≤ 4 3 .
Hence, f ( x , y , z ) ≤ 6 ( 4 3 ) 4 = 6 ∗ 2 8 3 4 = 2 7 3 5 .
Equality occurs when the values averaged are equal (to A say):
This would be when A = ( 1 − x ) = ( 1 − 2 y ) = ( 1 − 3 z ) = ( x + 2 y + 3 z ) = 3 − 3 A . Hence, A = 4 3 , x = 4 1 , y = 2 1 , z = 4 3 , these indeed satisfy the conditions in the question. Therefore, 2 7 3 5 is the maximum value.