Maximizing f(x,y,z) with Constraint and Lagrange Multipliers

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The discussion focuses on maximizing the function f(x,y,z) = x² + y² + z² under the constraint x⁴ + y⁴ + z⁴ = 1 using Lagrange multipliers. The initial setup of the Lagrangian is discussed, with attempts to solve for critical points leading to confusion about the maximum value. It is clarified that the maximum value of f is 1, achieved when one variable equals 1 and the others are zero, while the minimum value is √3. The importance of satisfying the constraint is emphasized, as incorrect values do not meet the requirement. The use of second-order tests for verifying maxima and minima is also mentioned.
BeBattey
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Homework Statement


Maximize f(x,y,z)=x^{2}+y^{2}+z^{2} with constraint x^{4}+y^{4}+z^{4}=1 using Lagrange multipliers

The Attempt at a Solution


I've got the setup as:
\Lambda(x,y,z,\lambda)=x^{2}+y^{2}+z^{2}+\lambdax^{4}+\lambday^{4}+\lambdaz^{4}+\lambda

I solve for all partials nice and clean, and spit out \sqrt{3} as the minimum value fine, with x=y=z=sqrt(6)/4 but I can't for the life of my find out how the max is 1.

Any and all help, even direction greatly appreciated.
 
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BeBattey said:

Homework Statement


Maximize f(x,y,z)=x^{2}+y^{2}+z^{2} with constraint x^{4}+y^{4}+z^{4}=1 using Lagrange multipliers

The Attempt at a Solution


I've got the setup as:
\Lambda(x,y,z,\lambda)=x^{2}+y^{2}+z^{2}+\lambdax^{4}+\lambday^{4}+\lambdaz^{4}+\lambda

I solve for all partials nice and clean, and spit out \sqrt{3} as the minimum value fine, with x=y=z=sqrt(6)/4 but I can't for the life of my find out how the max is 1.

Any and all help, even direction greatly appreciated.

Using the symbol u instead of \lambda, the Lagrangian for max or min f subject to g = 0 is L = f + u*g (or f - u*g---it just amounts to a sign change in u). Anyway, choosing the +u form the Lagrangian is x^2+y^2+z^2+u(x^4+y^4+z^4-1), so the optimality conditions are 2x + 4ux^3 = 0, etc. Either x = 0 or x = +-1/sqrt(-2u), etc. If we choose x = y = z = 1/sqrt(-2u) and impose the constraint, we have u = -sqrt(3)/2, giving x=y=z=1/3^(1/4) and f = sqrt(3). If we choose x = 0 and y = z = 1/sqrt(-2u), the constraint gives u=-1/sqrt(2), which gives x=0, y=z=1/2^(1/4) and f = sqrt(2). If we choose the solution x=y=0, z=1/sqrt(-2u) the constraint gives u=-1/2, giving z = 1 and so f = 1. Therefore, 1 is the minimum of f and sqrt(3) is the maximum. The other value f = sqrt(2) is at a "saddle point".

We could also use second-order tests to check for max/min, namely, to check whether the Hessian of the Lagrangian (not the function f) is positive definite, negative definite or indefinite at a given stationary point (x*,y*,z*,u*).

I verified my solution using the global solver in LINGO11, solving both the max and min problems.

RGV
 
Last edited:
Unfortunately, it is impossible to say what you did wrong because you do not show how you got that answer, but if you believe that x= y= z= \sqrt{6}/4 gives a max or min for this problem, you must be completely misunderstanding the problem because they do not satisfy the constraint:
x^4+ y^4+ z^4= \frac{36}{64}+ \frac{36}{64}+ \frac{36}{64}= \frac{9}{16}+ \frac{9}{16}+ \frac{9}{16}= \frac{27}{16}
NOT 1!
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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