Find the radical maximum value
3 x + 1 + 3 y + 3 + 3 z + 5
Suppose x , y , z are non-negative real numbers such that x + y + z = 1 , then find the maximum value of the expression above.
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The answer is 6.
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1 solution
Anik, the issue with your problem was that it stated "non-negative integers", which forced { x , y , z } = { 0 , 0 , 1 } . I've updated it to "non-negative real numbers", which allows for a larger set of possibilities.
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@Calvin Lin I think that, in order for the answer to be 6 , we would have to allow for x , y , z to be any real numbers. I used Lagrange multipliers, and found that the maximum is achieved when ( x , y , z ) = ( 1 , 3 1 , − 3 1 ) , which makes each of the terms equal to 2 . I can't see how to achieve the maximum over the set of non-negative real numbers.
That's fine!
You forgot to find values for which the maximum value can be achieved. Your answer achieves 6 at (x,y,z) = (1, 1/3 , -1/3), which is not a valid solution because z is negative. I believe that (x,y,z) = (5/6, 1/6, 0) has the greatest value for this problem, about 5.977.
Suppose 3 x + 1 = a 2 , 3 y + 3 = b 2 and 3 z + 5 = c 2
a 2 + b 2 + c 2 = 3 ( x + y + z ) + 9 = 1 2
Now, ( a + b + c ) 2 ≤ 3 ( a 2 + b 2 + c 2 )
( a + b + c ) 2 ≤ 3 . 1 2
( a + b + c ) ≤ 6
From this we get the maximum value of a + b + c = 6
Or 3 x + 1 + 3 y + 3 + 3 z + 5 is less than equal to 6 .