Solving a system in five unknowns for lagrange multiplier

[jonroberts74]

Homework Statement

I have to find the extrema of a given function with two constraints

[tex]f(x,y,z) = x+y+z;x^2-y^2=1;2x+z=1[/tex]

The Attempt at a Solution

If I create a new function F

then I have

[tex]F(x,y,z,\lambda,\mu)=x+y+z-(x^2\lambda - y^2\lambda -\lambda) -(2x\mu + z\mu - \mu) [/tex]

and taking the partials

[tex]\left\{\begin{array}{cc} F_{x} = 1-2x\lambda - 2\mu =0\\ F_{y} = 1+2y\lambda = 0 \\F_{z} = 1 - \mu = 0 \\ F_{\lambda} = -x^2 + y^2 + 1 = 0 \\ F_{\mu} = -2x - z +1 = 0 \end{array}\right. [/tex]

so now,

[tex]\mu =1[/tex]
solving for lambda

[tex]\lambda = -\frac{1}{2y} = \frac{1-2}{2x}[/tex]

now solving for x [or y] [tex]x=y[/tex]

but this causes an issue with [tex]-x^2+y^2+1=0[/tex] because 1 does not equal zero

 

[ehild]

There are no local extrema then.

ehild

 

[jonroberts74]

okay, that's what I had figured. Making sure Idid not miss something

 

[ehild]

But can be extrema on the boundary.

ehild

 

[jonroberts74]

But can be extrema on the boundary.

ehild

I can't find the zeros for the constraints though so what would I test in the function to see if there is.

 

[Ray Vickson]

But can be extrema on the boundary.

ehild

No. The fact that the Lagrangian equations plus the constraints have no solution precludes any extrema on the boundary.

This can also be verified directly: the two constraints can be used to find x and z in terms of y, then the results can be substituted into f(x,y,z); there are two solutions for x and z in terms of y, hence two versions of f. The two resulting problems are each one-dimensional problems, which can be examined graphically to check for the absence of local extrema. In other words, the "max" is +∞ and the "min" is -∞, but without any finite local constrained extrema at all.