Finding Max
Given that x , y , z are real numbers that satisfy the condition x 2 + y 2 + z 2 = 1 , find the maximum value of 2 x y + y z .
The answer is 0.866.
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8 solutions
Beautiful solution :) I kept thinking of a sphere looking at the constraint equation :D
Can you use AM-GM since the problem states that x,y,z are real but not necessarily positive? Although I used Lagrange multipliers and got the same result can you explain a bit more about justifying AM-GM here? Also is there a solution using CS inequality? I'll have to keep tinkering with this problem haha. Thanks for posting it!
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Yes we can, and we have to justify that step. Clearly, 2 x y + y z ≤ 2 ∣ x ∣ ∣ y ∣ + ∣ y ∣ ∣ z ∣ , which is why we can assume that we only care about the positive (non-negative) values.
Alternatively, we are applying AM-GM to the positive values of X = x 2 and Y = y 2 . Thus, we have x 2 + y 2 = X + Y ≥ 2 X Y = 2 ∣ x y ∣ ≥ 2 x y , the step is valid.
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Ok thanks for the explanation and clarification. This is really concise and clear.
I kept messing with the AM-GM solution and when you're verifying the equality case everything pretty much works out in the wash. The maximum is obtained when x,y,z all have the same sign whether positive or negative because of the xy and yz terms. Really interesting problem! I discounted AM-GM too soon when I saw x,y,z in real but not positive reals!
Awesome solution. Good job linking to the wiki too.
Wow. I kept using Lagrange Multipliers and wasted 14 minutes solving wherein it can be solved by less than a minute or 2 if I notice the factors behind the constraint equation. Good Job Mam =)
Can u please explain what you have done in step 2.
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@Harsh Kushwaha I have used the AM-GM inequality -- please see the link provided in the solution.
Interesting question confused me a lot, I think you should convert real numbers to positive real numbers in the question as this allows to apply AM-GM.
Although an elegant method has already been posted, I'd like to add my approach for the sake of variety!
By CS inequality and AM-GM we have, 2 x y + y z ≤ 3 × x 2 y 2 + y 2 z 2 = 3 × y 2 ( 1 − y 2 ) ≤ 3 2 1 = 2 3
using AM-GM on y 2 and 1 − y 2 .Equality occurs at x = 3 1 ; y = 2 1 ; z = 6 1
how did you get first inequality?
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Adarsh used Cauchy-Schwarz inequality: 2 x y + 1 y z ≤ [ 2 + 1 ] [ ( x y ) 2 + ( y z ) 2 ] . Sorry for not using LATEX. I will try to get around to learning it at some point.
it is a plain application of CS inequality!
Oh wow thank you so much for posting this! A very nice application of both inequalities
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No problem buddy!This is what we do!We teach and learn!
Substitute y = sin θ , x = sin α ∗ cos θ and z = cos α ∗ cos θ . The expression to be maximized becomes sin θ ∗ cos θ ∗ ( 2 sin α + cos α ) . Now the rest is easy.
And I thought I am the only one to use this method :D
How would you go about maximizing the trigonometric expression?
Could you give a more complete solution? I am very interested in your method, and I am sure others are as well.
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I would like to give some details discussion of preceeding trigonometric expression. Using y = sin θ , x = sin α ∗ cos θ , z = cos α ∗ c o s θ , where we see what is the meaning of α and θ ; using some imagination, that there is a 3-dimention space, composed of x-axis, y-axis, z-axis, and the substition formula means that θ is the angle of some strightline, which is connected with a point in the sphere and the origin, form a angle with the plane x-z, then α is the angle in plane x-z. After the substition, 2 x y + y z = 2 sin θ cos θ sin α + sin θ cos θ cos α = sin θ cos θ ( 2 sin α + cos α ) , using sin θ cos θ = 2 1 sin 2 θ ,and a c o s α + b sin α = a 2 + b 2 1 sin ( α + β ) ,where tan β = b a , then 2 x y + y z = 2 sin θ cos θ sin α + sin θ cos θ cos α = sin θ cos θ ( 2 sin α + cos α ) = 2 1 sin 2 θ ( 3 sin ( α + β ) = 2 3 sin 2 θ sin ( α + β ) ,where β is constant and tan β = 2 1 , so the maxmium value is depending on α and θ , and because they are independent with each other, with θ = 4 π and α + β = 2 π , namely, α = 2 π − β , there is a maxmium value that 2 3 sin 2 θ sin ( α + β ) ≤ 2 3 ∗ 1 ∗ 1 = 2 3
Using Lagrangian multipliers is an easy way to solve the problem.
From the constraint, we can see that x, y, z lie on a sphere. So we can use the parametric equation of a sphere with radius 1 instead of x, y, z and that makes the function sin(a) cos(a) (sqrt(2)*sin(b)+cos(b)) which has a max value of cos(30)
How would you go about maximizing the trigonometric expression?
The pattern is easily found -
Let a i { x , z } a n d b i { 2 y , y } Applying Cauchy Schwarz, we have -
( 2 x y + y z ) 2 ≤ ( x 2 + z 2 ) ( 3 y 2 )
( 2 x y + y z ) 2 ≤ ( 1 − y 2 ) ( 3 y 2 )
The RHS is a quadratic in y 2 with downward concavity and maxima at y 2 = 1 / 2
Substituting, we get , ( 2 x y + y z ) ≤ 2 3
P.S- I don't find it correct to use AM-GM since positive reals are not mentioned. In such cases generally, Cauchy Schwarz is the best Inequality. However, it may be trivial to prove that for the maximum case, the parameters must be positive. :)
I like your solution.
I played around with this problem and discovered that the maximum occurs at +/- (x.y.z). I don't remember the exact order of values for equality but they were like sqrt(1/2), sqrt(1/3), and sqrt(1/6) and the negative of this point. I'm in the same boat as you about being uncertain of using AM-GM here since they aren't positive reals. I'm sure it has something to do with symmetry so that it works out in this particular problem. At least in the inequality after applying CS we can set y^2 to some parameter which we know will be a positive real and AM-GM can be used.
It is incorrect to use AM-GM while considering x , y , z ∈ R , but we can consider only x , y , z ∈ R + since 2 x y + y z ≤ 2 ∣ x ∣ ∣ y ∣ + ∣ y ∣ ∣ z ∣ .
Taking f(x)=sqrt(2)xy+y sqrt(1-x^2-y^2) and using the partial derivatives, found we find the answer.
This is for the sake of variety.
We can substitute z in the expression as follows: z = 1 − x 2 − y 2 ⇒ 2 x y + y − x ² − y ² + 1
That new expression can be differentiated by x and y giving two equations both equal to 0. This gives three solutions, pick the one with the highest value.
Relevant wiki: Arithmetic Mean - Geometric Mean
1 = x 2 + y 2 + z 2 = ( x 2 + 3 2 y 2 ) + ( 3 1 y 2 + z 2 )
⩾ 2 3 2 x y + 2 3 1 y z = 3 2 ( 2 x y + y z ) ,
giving 2 x y + y z ⩽ 2 3 .