含多个约束的拉格朗日乘数法

Two constraints shrink the search space

A worked two-constraint problem

Reading the multipliers as prices

Distance problems are the natural showcase

With constraints $g_1 = c_1$ and $g_2 = c_2$, the Lagrange condition at a constrained extremum reads which way?

In the worked problem, $\lambda_1 = \tfrac{2}{3}$ and $\lambda_2 = \tfrac{14}{3}$. What does the comparison mean?

To find the point of a constrained set nearest the origin, which objective does the method prefer?

A Lagrange problem in $n = 3$ variables with two constraints has $m$ equations in $m$ unknowns. Give $m$, as a plain number.

The worked problem's nearest point $\left(\tfrac{8}{3}, \tfrac{1}{3}, \tfrac{7}{3}\right)$ has squared distance $\tfrac{38}{3}$ from the origin. Give the distance itself, in the form sqrt(a/b) like sqrt(38/3).