拉格朗日乘子：原点到平面的最短距离

最小化 $D^2=x^2+y^2+z^2$，约束 $2x+3y+4z=12$。

用拉格朗日乘子或直接套点到平面距离公式：

$$D=\frac{|12|}{\sqrt{2^2+3^2+4^2}}=\frac{12}{\sqrt{29}}.$$

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Original Explanation

Minimize $D^2=x^2+y^2+z^2$ subject to $2x+3y+4z=12.$
Construct $$\mathcal{L}(x,y,z,\lambda) = x^2 + y^2 + z^2 + \lambda\,(2x+3y+4z-12).$$ Setting partial derivatives to 0: $$\begin{cases} 2x + 2\lambda = 0,\\ 2y + 3\lambda = 0,\\ 2z + 4\lambda = 0,\\ 2x + 3y + 4z = 12. \end{cases}$$ Solve to get $x=\tfrac{24}{29},\,y=\tfrac{36}{29},\,z=\tfrac{48}{29},$ so $$D = \sqrt{x^2+y^2+z^2} = \frac{12}{\sqrt{29}}.$$

(Alternatively, use the plane-to-point distance formula $\,\tfrac{|\,d\,|}{\sqrt{\,a^2+b^2+c^2\,}}.$)