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Find the extrema of x^{2}+y^{2}+z^{2}\, subject to the constraint x^{2}+2y^{2}-z^{2}-1=0\,.

Using Lagrange multipliers, make a new function:


f=x^{2}+y^{2}+z^{2}+\lambda x^{2}+2\lambda y^{2}-\lambda z^{2}-\lambda =0\,

Take the partial derivatives of this function with respect to x, y, and z.

1) {\frac  {\partial f}{\partial x}}=2x+2\lambda x=0\,

2) {\frac  {\partial f}{\partial y}}=2y+4\lambda y=0\,

3) {\frac  {\partial f}{\partial z}}=2z-2\lambda z=0\,

For each equation above, find a value of lambda that works and force that value into the other two equations. Choose the other two variables appropriately to satisfy the equalities.

From 1), \lambda =-1, and in that case y=z=0. Plugging into our constraint equation yields x=1,-1.

From 2), \lambda =-1/2, and in that case x=z=0,y=1/{\sqrt  {2}},-1/{\sqrt  {2}}.

From 3), \lambda =1, and in that case x=y=0,z=i,-i.

Combining these results, the extrema are

(1,0,0),(-1,0,0),(0,1/{\sqrt  {2}},0),(0,-1/{\sqrt  {2}},0),(0,0,i),(0,0,-i)\,

The last two solutions may be disregarded if we are restricting to real variables.

Calculus of Variations

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