5

I'll expand on @QMechanic's answer. The first equation is straightforward. For the second, note your action is$$S=\int d^4x\mathcal{L},\,\mathcal{L}:=\frac12\rho v_iv_i-u(\rho)+\phi\dot{\rho}+\phi\rho\partial_iv_i+\phi v_i\partial_i\rho,$$so$$0=\frac{\partial\mathcal{L}}{\partial v_i}-\partial_j\frac{\partial\mathcal{L}}{\partial\partial_jv_i}=\rho v_i+\phi \...


4

Counter-example Here is a counter-example to show the solution will not necessarily be proportional to $|p(x)f(x)|$ over all $x \in E$, and the problem can have degenerate cases: Define: $E=[0,1]$ with the usual Lebesgue measure. $p(x) = 1, f(x)=x$ for all $x \in [0,1]$. For each $d \in (0,1]$ define $$q_d(x) = \left\{ \begin{array}{ll} \frac{2x}{d^2} ...


2

$\nabla f$ is the direction of greatest local increase in the function $f$. Moving perpendicular to $\nabla f$ causes no local change in the function. Thus, a local optimum on the constraint surface must occur when every direction you can move on the surface is perpendicular to $\nabla f$. If the equation of the constraint surface is given in the form $g(\...


2

The answer occurred to me as I was writing the question. But since I had already put a lot of work into the question, I decided to leave it there and answer it for my own reference. The error in my thinking was assuming that (7.8) would remain unchanged if the multiplier accompanying $||w||^2$ (the objective function) was changed.


2

Consider the first two equations, under this slightly different form:$$\left\{\begin{array}{l}(1+4\lambda)x+6y=0\\6x+(2+\lambda)y=0.\end{array}\right.$$Suppose that this homogeneous system has exatly one solution; then this solution is $(x,y)=(0,0)$, which is not a solution of the third one. But\begin{align}\text{The system has more than one solution}&\...


1

There's one extra constraint you've omitted—namely $$ \sum_a x_a = 1\ . $$ The Lagrangian function $\ \mathcal{L}\ $for the optimisation problem, then, is given by \begin{eqnarray} \mathcal{L}(x,\lambda) &=& \sum_a \left[x_a U(w,a) -\frac{x_a} {\beta}\log\frac{x_a}{p_0(a)}\right]-\lambda\left(\sum_ax_a-1\right)\\ &=& \lambda + \sum_a x_a\left[...


1

The problem can restated more clearly as looking for $$\text{argmax}_{x\in \mathbb R^n} \sum_{i=1}^n x_i U_i -\frac 1\beta x_i\log\frac{x_i}{y_i} \quad \text{such that }\sum_{i=1}^n x_i=1 $$ Note that the constraints $x_i\geq 0$ are implicit since $\log x_i$ appears in the objective. The Lagrangian is $$\mathcal L(x,\gamma) = \sum_{i=1}^n x_i U_i -\frac 1\...


1

The EL equations become: Variation wrt. $\phi$ $\qquad\Rightarrow \qquad \dot{\rho} + \nabla \cdot (\rho \mathbf{v})~=~0.$ Variation wrt. $\mathbf{v}$ $\qquad\Rightarrow \qquad \mathbf{v}~=~ \nabla\phi .$ Variation wrt. $\rho$ $\qquad\Rightarrow \qquad \frac{1}{2}|\mathbf{v}|^2 ~=~ u^{\prime}(\rho) + \dot{\phi} + \mathbf{v}\cdot \nabla\phi .$


1

Since AM $\geq$ GM, $\frac{x+y+z}{3} \geq (xyz)^{\frac{1}{3}}$ Therefore $(xyz)^{\frac{1}{3}} \leq \frac{1}{3}$ Alternatively you can do the following: Let $f(x, y, z) = (xyz)^{\frac{1}{3}} + \lambda (x+y+z -1)$ Set $\frac{\partial f}{\partial x} = \frac{\partial f}{\partial y} = \frac{\partial f}{\partial z} = 0$ Simplifying, $x = y = z = \frac{1}{...


1

One issue is going from: $a_{i}^{T}Xa_{i} \leq 1$ to: $\mbox{tr}((a_{i}a_{i}^{T})X) \leq 1$. This is done using a couple of standard tricks: A. Since $\mbox{tr}(s)=s$ for any scalar, $a_{i}^{T}Xa_{i} = \mbox{tr}(a_{i}^{T}Xa_{i}) \leq 1$. B. By the cyclic property of the trace of a product, $\mbox{tr}((a_{i}a_{i}^{T})X) \leq 1$. A second issue ...


1

Setting up the Lagrangian, \begin{align*} L = \sum_{i=1}^{n}x_i \log(x_i) + \lambda\left(\sum_{i=1}^{n}c_i x_i - m\right) + \mu\left(\sum_{i=1}^{n}x_i - 1\right) \end{align*} Computing the stationary points, \begin{align*} \frac{\partial L}{\partial x_i} = \log(x_i) + 1 + \lambda c_i + \mu\overset{\text{set}}{=} 0 \implies x_i = e^{-1 - \lambda c_i - \mu} \...


1

This is easy to implement and numerically solve in CVXPY, using the supplied function entr . The entropy maximization example at https://www.cvxpy.org/examples/applications/max_entropy.html?highlight=entropy can be modified for your problem by changing the constraints.


1

This "solution" is rather a story about searching for the solution in this case and in similar cases. It follows standard ideas to solve such inequalities. (I decided to wait till the one or the other one answer of Michael Rozenberg was accepted, then post the thoughts towards a solution, based on a paperless discussion, where i could not show the details.)...


1

There is also the following way: $$\frac{x}{y}+\frac{y}{z}+\frac{z}{x}-x^2-y^2-z^2=\frac{x}{y}+\frac{y}{z}+\frac{z}{x}-\frac{9(x^2+y^2+z^2)}{(x+y+z)^2}=$$ $$=\frac{\sum\limits_{cyc}x(x-y)^2(2y^2-7yz+8z^2)}{2xyz(x+y+z)^2}\geq0.$$ We can get it by the following play. Let $$\frac{x}{y}+\frac{y}{z}+\frac{z}{x}-\frac{9(x^2+y^2+z^2)}{(x+y+z)^2}=\frac{\sum\limits_{...


1

Clearly, none of the constraints $p_k > 0$ will be active at an optimal solution and the constraint $\sum_j p_j \leq 1$ will be active at an optimal solution. To see this, note that the point $p = (\frac{1}{K},\cdots,\frac{1}{K})$ is feasible for the true problem, and any feasible solution that has a component $p_j$ that is too small will have worse ...


1

Here is a problem that can be solved with solution proportional to $|p(x)f(x)|$. Problem Given: $(E, \mathcal{E}, \lambda)$ Measurable functions $p:E\rightarrow [0,\infty)$, $f:E\rightarrow\mathbb{R}$ $\int_E p(x)d\lambda = 1$ $0< \int_E |f(x)p(x)|d\lambda < \infty$. $p(x)f(x)\neq 0$ for all $x \in E$. We want to find a measurable function $q:E\...


1

We are given that $f(x,y,z)= x^2+y^2+z^2$ with the constraint $x + 2y+3z=15$. Let's define $g(x,y,z)=x + 2y+3z - 15$ and then relate partial derivatives between $f$ and $g$. We will multiply all the partial derivatives of $g$ by our Lagrange multiplier $\lambda$. $$2x=\lambda$$ $$2y=2\lambda$$ $$2z=3\lambda$$ We now need to solve all three of these ...


1

by cauchy Schwarz we get $$(x+2y+3z)^2\le (x^2+y^2+z^2)(1+4+9)$$ so $$\frac{225}{14}\le x^2+y^2+z^2$$


1

I think by C-S is better: $$15=\frac{(2x+3y+5z)\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{5}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{5}}\geq\frac{(\sqrt{x}+\sqrt{y}+\sqrt{z})^2}{\frac{1}{2}+\frac{1}{3}+\frac{1}{5}}=\frac{(\sqrt{x}+\sqrt{y}+\sqrt{z})^2}{\frac{31}{30}},$$ which gives $$\sqrt{x}+\sqrt{y}+\sqrt{z}\leq\sqrt{15.5}.$$ The equality occurs for $$\left(\...


1

I will give you the actual solution only so that you may verify your answer at the end. If this is a homework problem, you would be committing some pretty egregious academic misconduct if you just took this answer and ran with it. Spoiler: The solution is $x=\frac{225}{62}, y=\frac{50}{31}, z=\frac{18}{31}$. If you had a copy of Wolfram Mathematica ...


1

$1)$ Let $F(x,y,z,\lambda) \equiv F= \sqrt x+\sqrt y+\sqrt z + \lambda(2x+3y+5z-15)$ $2)$ Compute $F_x = F_y = F_z = F_\lambda = 0$ $3)$ Find $x,y,z$ in terms of $\lambda$ $4)$ Use $x,y,z$ (in terms of $\lambda$) in $2x+3y+5z =15$ and find the value(s) of $\lambda$ and hence those of $x,y,z$. $5)$ From the set(s) of values of $x,y,z$ find which one ...


1

Hint: The exponent law is $\left(a^b\right)^c=a^{b\cdot c}$. But you don´t have brackets here. Therefore the first partial derivative w.r.t. $k$ is $1-\lambda \cdot 3\cdot A^{1/4}\cdot k^{A^{1/4}-1}\cdot h^{A^{h/4}}=0\Rightarrow 1=\lambda \cdot 3\cdot A^{1/4}\cdot k^{A^{1/4}-1}\cdot h^{A^{3/4}}$. The first partial derivative w.r.t. $h$ is $1-3\cdot \...


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