This question already has an answer here: Global maximum and minimum of $f(x,y,z)=xyz$ with the constraint $x^2+2y^2+3z^2=6$ with Lagrange multipliers? 2 answers Find the Lagrange multipliers with one constraint: $f(x,y,z) = xyz$ and $g(x,y,z) = x^2+2y^2+3z^2 = 6$ 1 answer

Show that the volume of the largest rectangular parallelepiped that can be inscribed in the ellipsoid $$\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1$$ is $\dfrac{8abc}{3\sqrt3}$. I proceeded by assuming that the volume is $xyz$ and used a Lagrange multiplier to start with $$xyz+\lambda \left(\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}-1\right)$$ I proceeded further to arrive at $\frac{abc}{3\sqrt3}$. Somehow I seemed to be have missed $8$. Can someone […]

Optimize $f(x,y,z) = 4x^2 + 3y^2 + 5z^2$ over $g(x,y,z) = xy + 2yz + 3xz = 6$ According to the theorem the gradients must be parallell, $\nabla f = \lambda \nabla g$, so their cross product must equal zero i.e. $\nabla f \times \nabla g = \mathbf{0}$. This results in the following system of […]

We define $f:\mathbb{R}^n\times\mathbb{R}^n\to\mathbb{R}$ by $f(u,v)=\displaystyle\sum_{i=1}^n|u_i-v_i|^p$. We’d like to minimize $f$ under the following constraints: $$ \left\{\begin{array}{l} g_1(u,v):=f(u,0)=1 \\ g_2(u,v):=f(0,v)=1 \\ g_3(u,v):=\langle u,v\rangle=0\end{array}\right. . $$ I know one way is to use Langrage multipliers to find the local maximum and minimum, so I restricted $f$ to the set $$D=\{(u_1,…,u_n,v_1,…,v_n):u_i\neq0,v_i\neq0,u_i\neq v_i,\forall 1\leq i\leq n\},$$ tried to solve […]

If $f(x,y,z,\ldots)$ is symmetric in all variables, (i.e $f$ remains the same after interchanging any two variables), and we want to find the extrema of $f$ given a symmetric constraint $g(x,y,z,\ldots)=0$, $$\bf\text{When is it true that the extrema is achieved when }\ x=y=z=\ldots?$$ An example where this claim is true: $$ g(x,y,z) = x+y+z – […]

Given 2 vectors $u,v \in \mathbb{R^n}$ such that $\|u\| = 1$ and $\sum_{i=1}^n v_i= c$ where $c<1$, I would like to maximize $$\sum_{i=1}^n u_i v_i \log (v_i)$$ and minimize $$\sum_{i=1}^n u_i v_i \log (u_i)$$. The answer in my opinion is $-c\log(c)$ if $c<1/e$ (and $c\log(c)$ for the minimum). I think so because on $[-1/e, 1/e]$ […]

Given a functional $$J(y)=\int_a^b F(x,y,y’)dx, \tag{1}$$ where $y$ is a function of $x$, and a constraint $$\int_a^b K(x,y,y’)dx=l, \tag{2}$$ if $y=y(x)$ is an extreme of (1) under the constraint (2), then there exists a constant $\lambda$ such that $y=y(x)$ is also an extreme of the functional $$\int_a^b [F(x,y,y’)+\lambda K(x,y,y’)]dx. \tag{3}$$ Similarly, if the constraint is […]

I am reading the following tutorial on Lagrangian multipliers (http://www.cs.berkeley.edu/~klein/papers/lagrange-multipliers.pdf). My goal is to gain an intuitive understanding of why the Lagrangian method works. In summary, the tutorial link above and other tutorials have helped me understand that constraint optimization boils down to having parallel normal vectors for the function to be maximized and the […]

This is related to two previous questions which I asked about the history of Lagrange Multipliers and intuition behind the gradient giving the direction of steepest ascent. I am wondering if the constant $\lambda$ in the Lagrange equation $$\nabla f=\lambda \nabla g$$ has any significance. For instance, can the sign of $\lambda$ tell us anything […]

We know that there are two definitions to describe lasso. Regression with constraint definition: $$\min\limits_{\beta} \|y-X\beta\|^2, \sum\limits_{p}|\beta_p|\leq t, \exists t $$ Regression with penalty definition: $$\min\limits_{\beta} \|y-X\beta\|^2+\lambda\sum\limits_{p}|\beta_p|, \exists\lambda$$ But how to convince these two definition are equivalent for some $t$ and $\lambda$? I think Lagrange multipliers is the key to show the relationship between two […]

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