Solving a system in five unknowns for lagrange multiplier

1. Jul 12, 2014

jonroberts74

1. The problem statement, all variables and given/known data

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

$$f(x,y,z) = x+y+z;x^2-y^2=1;2x+z=1$$

3. The attempt at a solution
If I create a new function F

then I have

$$F(x,y,z,\lambda,\mu)=x+y+z-(x^2\lambda - y^2\lambda -\lambda) -(2x\mu + z\mu - \mu)$$

and taking the partials

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

so now,

$$\mu =1$$
solving for lambda

$$\lambda = -\frac{1}{2y} = \frac{1-2}{2x}$$

now solving for x [or y] $$x=y$$

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

2. Jul 13, 2014

ehild

There are no local extrema then.

ehild

3. Jul 13, 2014

jonroberts74

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

4. Jul 13, 2014

ehild

But can be extrema on the boundary.

ehild

5. Jul 13, 2014

jonroberts74

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

6. Jul 13, 2014

Ray Vickson

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.

Last edited: Jul 13, 2014