Not 64/243
P = x 5 y + y 5 z + z 5 x
Let x , y , z be non-negative real numbers such that: x + y + z = 2 .
Find the maximum value of the expression above.
The answer is 4.287.
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1 solution
Moderator note:
You should point out that "Lagrange multipliers work best for finding interior points that are max/min. But we also have to check the boundary, which occurs when (at least) one of the variables is 0.
It is for this reason, that most Lagrange multiplier solutions submitted in competitions, is considered incomplete unless the boundary conditions are properly checked.
I don't mean to ruin the fun but I don't think problem is posted in the spirit of a Lagrange Multiplier solution.
@Khang Nguyen Thanh Do you have a strictly algebraic solution?
Why should we consider the case where one of the variables is zero?
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By the extreme value theorem , the continuous function P does have a global maximum on the compact space x + y + z = 2 with non-negative x , y , z , but not necessarily on x + y + z = 2 with positive variables. Just as in single-variable calculus, we need to consider the "end points."
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Sir,could you please elaborate?
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@Tobi Major – Our constraint space, a triangular surface with boundary, is compact (closed and bounded), so that the extreme value theorem applies. By contrast, the space x + y + z = 2 with positive variables fails to be closed, and, indeed, the function P attains neither a global maximum nor a global minimum there.
The problems states that x , y , z are non-negative, so, we certainly need to consider the case when (exactly) one of them is zero. (If two of them are zero then P = 0 , the minimum of P )
Here is an analogy: When finding the extrema of a differentiable function on a compact interval [ a , b ] in single-variable calculus, we need to examine the critical points AND the endpoints too.
If x , y and z are all positive, then Lagrange multipliers give x = y = z = 3 2 , where the value is P = 2 4 3 6 4 . But we are told that this isn't the answer.
Now consider the case where exactly one of the variables is zero, for example, z = 0 . Then P simplifies to P = x 5 y and the maximum is attained when x = 5 y , by Lagrange multipliers, so that the maximum is 3 6 5 5 ≈ 4 . 2 8 6 , attained at ( x , y , z ) = ( 5 / 3 , 1 / 3 , 0 ) and two other points.