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

[BeBattey]

Homework Statement

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

The Attempt at a Solution

I've got the setup as:
[tex]\Lambda[/tex](x,y,z,[tex]\lambda[/tex])=x[tex]^{2}[/tex]+y[tex]^{2}[/tex]+z[tex]^{2}[/tex]+[tex]\lambda[/tex]x[tex]^{4}[/tex]+[tex]\lambda[/tex]y[tex]^{4}[/tex]+[tex]\lambda[/tex]z[tex]^{4}[/tex]+[tex]\lambda[/tex]

I solve for all partials nice and clean, and spit out [tex]\sqrt{3}[/tex] 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.

 

[Ray Vickson]

Homework Statement

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

The Attempt at a Solution

I've got the setup as:
[tex]\Lambda[/tex](x,y,z,[tex]\lambda[/tex])=x[tex]^{2}[/tex]+y[tex]^{2}[/tex]+z[tex]^{2}[/tex]+[tex]\lambda[/tex]x[tex]^{4}[/tex]+[tex]\lambda[/tex]y[tex]^{4}[/tex]+[tex]\lambda[/tex]z[tex]^{4}[/tex]+[tex]\lambda[/tex]

I solve for all partials nice and clean, and spit out [tex]\sqrt{3}[/tex] 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 [tex]\lambda[/tex], 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 [tex]x^2+y^2+z^2+u(x^4+y^4+z^4-1)[/tex], so the optimality conditions are [tex] 2x + 4ux^3 = 0[/tex], 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

 

[HallsofIvy]

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 [itex]x= y= z= \sqrt{6}/4[/itex] gives a max or min for this problem, you must be completely misunderstanding the problem because they do not satisfy the constraint:
[tex]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}[/tex]
NOT 1!