CoV2

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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), λ = − 1, and in that case y = z = 0. Plugging into our constraint equation yields x = 1, − 1.

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

From 3), λ = 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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