Max and min part 3
x 2 + y 2 + z 2
If 0 ≤ x , y , z ≤ 2 are number satisfying x + y + z = 3 , find the product of the maximum and minimum value of the above expression.
- Part of the max and min
The answer is 15.
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1 solution
Sir,in what type of inequality problems can this assumption of x=min{x,y,z} come handy?
Why must x = 0 when the maximum is attained?
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Actually, that's the part I felt really compulsive after finishing the solution, but I haven't found any prove to why one of the variables must equal to 0 yet
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I think this might be an argument :
The higher the positive number (larger than one), the more it's square gains. By this I mean that when you square 1, it stays one, so the square gains nothing. When you square two, it becomes 4, so it 'gains' 2. The quadratic function rises stronger for higher positive numbers.
With positive numbers between 0 and 1, there is even a loss.
To formalize the idea :
Assume (x,y,z) = (2,1,0) and let us assume that this couple of numbers doesn't give the greatest value in the expression.
There can be two cases : we get either two numbers between 1 and 2 and one between 0 and 2, or one between 1 and 2 and two between 0 and 1.
I will prove the first case and leave the second as an exercise (it's completely similar) .
Now suppose there exist a,b,c such that (x',y',z') = (2-a,1+b,c) with a = b + c such that these numbers maximize the expression. In the first case we know that a ≤ 2 1
Then
( 2 − a ) 2 + ( 1 + b ) 2 + c 2 > 5
4 − 4 a + a 2 + 1 + 2 b + b 2 + c 2 > 5
a 2 − 4 a + b 2 + 2 b + c 2 > 0
( b + c ) 2 − 4 ( b + c ) + b 2 + 2 b + c 2 > 0
2 b 2 + 2 b c + 2 c 2 − 2 b − 4 c > 0
b 2 + b c + c 2 > b + 2 c
( b + c ) 2 > b + c + c + b c
and this gives a contradiction because b + c ≥ ( b + c ) 2 .
Thus our assumption is false and (x,y,z) gives a larger value than (x',y',z').
Is the maximum of 9 when ( x , y , z ) = ( 0 , 0 , 3 ) possible?
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Look at the condition carefully, x , y , z ∈ [ 0 ; 2 ] , so that solution is invalid
Im pretty sure this can be solved with lagrange multiplier easiley
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I found the minimum easily with lagrange multipliers (it revealed right away that x,y and z were equal), but I wasn't sure how to get the max with that method.
Using Cauchy-Schwarz Inequality we have x 2 + y 2 + z 2 ≥ 3 ( x + y + z ) 2 = 3 The equality holds when x = y = z = 1
To find the maximum, we can apply Oxyz. The point A ( x , y , z ) satisfies the constraint lies on the hexagon created by the plane x + y + z − 3 = 0 and the cube with length 2. Consider the hexagonal created by it and have O as it apex, B ( 1 , 1 , 1 ) is the feet of the altitude descend from O and also the centerpoint of the hexagon. Each hypotenuse of a triangle created by O, B and one of the hexagon edges is the farthest distance from O to any point on the hexagon. Calculation result in coordinates of those edges being permutations of ( 1 , 2 , 0 ) , we'll find out the square of those hypotenuse is 5 in length So the product is 3 . 5 = 1 5