What Is The Maximum?
If x , y , z are real numbers such that x + y + z = 3 , what is the maximum value of
2 x + 1 3 + 3 3 y + 5 + 4 8 z + 1 2 ?
This question was posed in SMO 2012. I came across it and it was interesting.
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2 solutions
Hm, good solution. If you like hard problems, invite you to try one of mine; they're from a very old math book.
Cheers
I didnt get it?? Can u please explain?
i did it the same way
If we set x = k − 2 1 3 , y = − k − 2 5 , z = 3 + 2 1 3 + 2 5 , then as k → ∞ , the expression looks like 2 k + 3 − 3 k , which tends to ∞ (at a pretty slow rate).
They forgot to add the condition that the terms should be non-negative. With this additional restriction, what is the maximum value of the expression? Justify your answer.
Maximum value when the nos. are non-negative will be 8, when x=3/2, y=1 and z=1/2
Vineeth Chelur how did u do it ?
i did the same way
Notice that O ( 2 x + 1 3 ) > O ( 3 3 x + 5 ) > O ( 4 8 x + 1 2 ) and O ( 3 3 x + 5 2 x + 1 3 ) = ∞ . Hence, that leads to setting z = 0 and finding that lim x → ∞ 2 x + 1 3 + 3 3 ( 3 − x ) + 5 = ∞ . There are no domain restrictions in f ( x ) = 3 x .