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proof of Lagrange multiplier method
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(Proof)
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Let $g(x,y)=c$ and $f(x,y)=d$ Taking the derivative of $f$ and $g$ with respect to $t$ gives:
$\displaystyle\frac{\partial f}{\partial t}=\frac{\partial f}{\partial x}x'(t)+\frac{\partial f}{\partial y}y'(t)=0$ endcenter
$\displaystyle\frac{\partial g}{\partial t}=\frac{\partial g}{\partial x}x'(t)+\frac{\partial g}{\partial y}y'(t)=0$ endcenter
By letting $\vec{r}=x(t)\hat{i}+y(t)\hat{j},$ the partial derivatives can be rewritten as follows:
$\displaystyle\frac{\partial f}{\partial t}=\operatorname{grad} f\cdot \vec{r'};$ $\displaystyle\frac{\partial g}{\partial t}=\operatorname{grad} g\cdot \vec{r'}$ endcenter
This implies that $\operatorname{grad} f \times \operatorname{grad} g = 0,$ thus $\operatorname{grad} f = \lambda \operatorname{grad} g.$ Now this equation can be rewritten as $f_x\hat{i}+f_y\hat{j}=\lambda\left(g_x\hat{i}+g_y\hat{j}\right).$ Since $\mathbb{R}^n\mapsto \mathbb{R},$ this equation can be separated into two new equations:
$\displaystyle f_x=\lambda g_x;\quad f_y=\lambda g_y$ endcenter
Using the above equations, a new function, $F$ can be defined:
$\displaystyle F(x,y,\lambda)=f(x,y)-\lambda g(x,y)$ endcenter
which can be generalized as:
$\displaystyle F(x,y,\lambda)=f(x,y)-\sum_{i=1}^m \lambda_i\left[g_i(x,y)\right].$ endcenter
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"proof of Lagrange multiplier method" is owned by aplant. [ owner history (2) ]
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Cross-references: function, separated, equation, implies, partial derivatives, derivative
This is version 3 of proof of Lagrange multiplier method, born on 2005-07-25, modified 2005-07-27.
Object id is 7263, canonical name is ProofOfLagrangeMultiplierMethod2.
Accessed 4255 times total.
Classification:
| AMS MSC: | 15A18 (Linear and multilinear algebra; matrix theory :: Eigenvalues, singular values, and eigenvectors) | | | 15A42 (Linear and multilinear algebra; matrix theory :: Inequalities involving eigenvalues and eigenvectors) | | | 45C05 (Integral equations :: Eigenvalue problems) |
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Pending Errata and Addenda
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