Irrelevant Additional Information?
{ w 2 + x 2 + y 2 + z 2 = 2 0 w x y z = 2 4
If w , x , y , z are 4 real numbers such that the system of equations above is true, what is the maximum value of x 2 + y 2 ?
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2 solutions
Moderator note:
Yes, Lagrange multipliers offer a simple and powerful technique to solve inequalities, especially when the terms are in polynomial form.
There is a solution that just uses quadratic equations and stating the obvious AM-GM relations :)
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I like Lagrange multipliers since they are such a systematic tool; one does not have to look for clever moves.
Here is an alternative method: Write x 2 + y 2 = 1 0 + d and z 2 + w 2 = 1 0 − d ; we wish to maximize d . Now ( x 2 + y 2 ) ( z 2 + w 2 ) = 1 0 0 − d 2 ≥ 4 x y z w = 9 6 . Thus d 2 ≤ 4 and d ≤ 2 . The maximum value of x 2 + y 2 is 1 2 , attained when x = y = 6 and z = w = 2 .
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Right, Lagrange multipliers is the "lazy man" way to deal with inequalities. It's a great tool once you mastered it (and most people always forget to check the boundary conditions).
Nice alternative method! That's what I did too, but with more symbols. I'm amazed that the information is just condensed into "d".
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@Calvin Lin – Sir, I tried to understand Lagrange multipliers but I could not understand it......Please can you help me.
I did it using Am -Gm relationship .
[This is not a complete solution.]
Now, you might be thinking that x 2 + y 2 ≤ w 2 + x 2 + y 2 + z 2 = 2 0 , and we could have equality if w = z = 0 . However, the issue is that it would not satisfy w x y z = 2 4 , and so this isn't viable. Neither can we put w , z → 0 in the hopes of getting w x y z = 2 4 , because we are subject to x 2 + y 2 ≤ 2 0 .
So, how do we proceed and make use of
w
x
y
z
=
2
4
? That will be left for you to figure out.
Hint:
x
2
+
y
2
≥
2
x
y
,
w
2
+
z
2
≥
2
w
z
.
Note: The maximum is achieved at x = y = 6 , w = z = 2 .
My favourite technique of Lagrange multipliers quickly shows that w 2 = z 2 ; this follows from the conditions 2 λ z + μ x y w = 0 and 2 λ w + μ x y z = 0 . Likewise, the other two equations give x 2 = y 2 . Now the constraints simplify to x 2 + z 2 = 1 0 and x 2 z 2 = 2 4 , with the two solutions x 2 = 6 , 4 . The maximum we seek is x 2 + y 2 = 2 x 2 = 1 2 (and the minimum is 8).
I know that Lagrange multipliers are frowned upon by some, but they do offer a simple and powerful technique that makes problems like this quite routine.