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2

I do not see the Lagrange multipliers of the tag being used, where is the $\lambda$? It is obviously best to use the full $274$, so we maximize $lwh$ subject to $1+2w+2h=274$. The Lagrangian is $lwh-\lambda(l+2w+2h)$. Set the partial derivatives equal to $0$. We get $wh-\lambda=0$, $lh-2\lambda=0$, $lw-2\lambda=0$. We have the additional equation ...


0

Let the total horizontal distance travelled be $a$. The light travels in the first medium a distance $d_{1}$ at an angle $\theta _{1}$ to the vertical. Then if the horizontal distance traveled in this step is $x$ and the vertical distance traveled is $h$, the time it takes to reach the border is $t_{1}=\frac{\left ( x^{2}+h^{2} \right )^{1/2}}{v_{1}}$. ...


2

Without Lagrange multipliers You don't need to use Lagrange multipliers. Snell's law can be derived directly from Fermat's principle of least time, which consists in minimizing the time taken by the ray of light to travel from one point to another. Take a look at the diagram below: In the first medium, with refractive index $n_1$, light will travel at ...


0

Hint: Use Fermat's principle. the path taken between two points by a ray of light is the path that can be traversed in the least time. And you don't need Lagrange multipliers.


0

The total travel time, which is the sum of travel times from $A$ to the surface and fromthe surface to $B$. The two part times are obtained as length divided by speed, the two lengths depend on the angles.


2

So, we have $V=\pi r^2 h=1$ and wish to minimize the surface area $S=2\pi r^2+2\pi rh$. HINT: Note that we have $V=\pi r^2 h=1$ as a constraint. So, eliminate $h$ as $h=1/(\pi r^2)$. Then, substitute this expression for $h$ into the expression for $S=2\pi r^2 + 2\pi r h$, take the derivative with respect to $r$, set this equal to zero and find the ...


2

It is a closed region, so max and min must occur. They can only occur on the boundary or at critical points of the function. So you can use the following steps: Step 1: Find all the critical points of the function, and check whether they are in the constraint region. Step 2: Use regular Lagrange multiplier method on the boundary of the disk. Then ...


1

Let $a=\frac{x}{x+y},b=\frac{y}{x+y}$. The inequality $$(\frac{x+y}2)^n \leq \frac{x^n+y^n}2$$ is equivalent to $$\frac{a^n+b^n}2\ge\frac{1}{2^n} $$ for $a,b>0$ and $a+b=1$. Let $$ f(a,b,\lambda)=\frac{a^n+b^n}2-\lambda[(\frac{a+b}2)^n-1]. $$ Then solving $$ \frac{\partial f}{\partial a}=\frac n2a^{n-1}-\frac{\lambda n}{2^n}(a+b)^{n-1}=0,\frac{\partial ...


1

Hint : Calculate the maximum of $(\frac{x+y}{2})^n$ and the minimum of $\frac{x^n+y^n}{2}$ with Lagrange multipliers and compare the results.


1

As you have already noticed, if $(x, y) \in g^{(-1)}\{ 0 \}$ such that $f$ has an extremum at $(x, y)$, then, since $\lambda = 0$ is easily seen to lead to a contradiction, we have $x = y.$ However, since $(x, y) \in g^{(-1)}\{ 0 \},$ we have $x^{2} - y^{2} = 2,$ which implies that $x \neq y.$ Therefore, $f$ restricted on $g^{(-1)}\{ 0 \}$ has no extremum. ...


0

One way to formulate the problem using ADMM is to let the ADMM-variable $X$ contain $A_x$ and $A_y$, i.e. $X = [A_x; A_y]$ (semi-colon denotes stacking, as in Matlab etc.), and let $Z=[Z_1; Z_2; Z_3; Z_4]$ contain four blocks corresponding to $A_x$, $A_x$, $A_x$ and $A_y$ respectively. (I will write $Q$ for $X-I$, where $X$ is your variable $X$! Thus $A_x ...


1

The expression $I[y] = \int_a^b \! mgy \, \mathrm{d}x$ is not quite correct. For one thing, $m$ should be replaced by linear density $\rho$, measured in mass per unit of length. Indeed, to get the mass of a short piece of the string, we multiply its length by $\rho$. The length is represented by $\sqrt{1+(y'(x))^2} \, \mathrm{d}x$. Thus, the correct ...


1

I don't think there is a very simple method to get the answer. But you can simplify the cases: case 1 : $C=0\cap A*B \neq 0 \implies y=-\dfrac{Ax}{B} $ , the rest should be easy. case2: $C\neq 0 \implies z= -\dfrac{Ax+By}{C} \implies (1+\dfrac{A^2}{C^2})x^2+(1+\dfrac{B^2}{C^2})y^2+2\dfrac{AB}{C^2}xy=1 \\f=\dfrac{1}{c^2}-(\dfrac{x^2}{p^2}+\dfrac{y^2}{q^2}), ...


1

I assume you mean the constraint $x^T Px + 2c^Tx + s \leq 0$. For intuition on the difficulty of this constraint, let us assume we also have constraints $x_i \in [0,1]$ for all $i \in\{1, \ldots, n\}$. Now consider your single constraint in the special case $P=-I$, $c=(1/2, \ldots, 1/2)$, $s=0$: $$ -x^Tx + 1^Tx \leq 0 $$ This is equivalent to saying: ...


0

As you said, your optimization problem can be reformulated as \begin{align*} \text{minimize} & \quad \sum_{i=1}^n a_i \xi_i + b_i \psi_i \\ \text{subject to} & \quad \xi_i \geq 0, \quad i = 1,\ldots, n \\ & \quad \xi_i \geq 1 - y_i (\beta_0 + \beta^T x_i) , \quad i = 1,\ldots, n\\ & \quad \psi_i \geq 0, \quad i = 1,\ldots, n \\ & \quad ...


0

Your function can be rewritten as $$f(x) = \begin{cases} a (1-x) & x<-1 \\ a(1-x)+b(1+x) & -1 \leq x \leq 1 \\ b(1+x) & x < 1\end{cases}.$$ The first and third case have no critical points, so the only possible local extrema from there are on the boundary: $f(-1)=2a$,$f(1)=2b$. There could also be a local extremum at a critical point in ...


-2

This is mechanized in Maple. with(Student[MultivariateCalculus]): LagrangeMultipliers(x*y*z, [x^2+2*y^2+3*z^2-6], [x, y, z], output = detailed); $$[x=0,y=0,z=\sqrt {2},\lambda_{{1}}=0,xyz=0],\,[x=0,y=0,z=\sqrt {2},\lambda_{{1}}=0,xyz=0],\,[x=0,y=\sqrt {3},z=0,\lambda_{{1}}=0,xyz=0],\,[x=0,y=-\sqrt {3},z=0,\lambda_{{1}}=0,xyz=0],\,[x=\sqrt ...


1

You should try to solve the equations. The six solutions wouldn't help. Following @Mann's idea, you can multiply the left hand sides of the three equations, and multiply the right hand sides of the three equations, then equate them: $$xy\cdot yz\cdot xz =48 \lambda^3 xyz$$ then see if you can solve the $x,y,z$'s with the obtained information. To make the ...


1

For a convex optimization problem, the KKT conditions are sufficient for a point to be a global minimizer. And your optimization problem is convex when $f$ is convex and $g_1$ and $g_2$ are affine. Consider the convex optimization problem \begin{align*} \text{minimize} & \quad f(x) \\ \text{subject to} & \quad Ax = b. \end{align*} Here $f:\mathbb ...


0

Easier: cross multiply the two gradients using the determinant method and equate the result with 0. You get xy=0=yz=xz. The candidates are (x,0,0),(0,y,0) and (0,0,z). Plugging it into the constraint you get (3,0,0),(-3,0,0),(0,6,0),(0,-6,0),(0,2,0),(0,0,-2). The last two give min, and the the middle two give max or check.


0

The Lagrangian tells you that there is no interior critical point unless all the $a_i$ are equal. Thus, in general, the minimum will be on the boundary. In fact, it is in a corner. Without loss of generality, let $a_0=\max\limits_{0\le i\le N-1}a_i$. Then $$ \begin{align} -\sum_{i=0}^{N-1}a_ib_i &=-a_0\sum_{i=0}^{N-1}b_i+\sum_{i=0}^{N-1}(a_0-a_i)b_i\\ ...


2

In 1D calculus, an example of a function $f : \mathbb R \to \mathbb R$ where $f'(x) = 0$ does not give a local max or min is $f(x) = x^3$. Inspired by that, try this example: Let $g(x,y) = x^2y$ with the constraint $h(x,y) = x - y = 0$. Then $\nabla g - \lambda \nabla h = 0$ iff $$2xy - \lambda = 0 \ \text{ and } x^2 + \lambda = 0$$ I.e., $2xy = -x^2$ or ...


0

I do believe your book answer is wrong, or the question has a typo. Your set up is correct. From my habit, I used $-\lambda$ in the following, but it should not give any difference: \begin{align*} 2x -\lambda (8x + 5y) &= 0\\ 2y - \lambda (5x + 6y) &= 0\\ 4x^2 + 5xy + 3y^2 - 9 &= 0 \end{align*} So from 1, $\lambda = \frac{2x}{8x+5y}$. Plug ...


0

I haven't worked out a numerical answer by hand: if I eliminate k from: $$\begin{align*} 2x + k(8x + 5y) &= 0\\ 2y + k(5x + 6y) &= 0\\ \end{align*}$$ I get: $$\begin{align*} 2y - \frac{2x(5x+6y)}{8x+5y} &= 0\\ \end{align*}$$ And if one substitutes in $$\begin{align*} x=\frac{3}{\sqrt{2}}\\ y=\frac{-3}{\sqrt{2}} \end{align*}$$ One would ...


0

Since your condition is an integral condition, and not pointwise along the path, the Lagrange multiplier $λ$ is a constant, which should somehow reduce the equations. In the end, you obtain solution curves that depend on $λ$, as does in consequence the value of the condition integral. Now you have a one-parameter problem $h(λ)=2$ to solve, where the ...


0

The "constraint" $Ax+Bz=c$ is there to introduce variable splitting. The actual (problem) constraints are introduced through indicator functions (e.g. $g(z) = \mathbf{1}_{\|\cdot\| \leq \varepsilon} (z)$). To allow for multiple constraints, you can reformulate your problem such that $f$ or $g$ represents a sum of penalty functions (approach known as SDMM, ...


2

Here is a moderately complicated example, cooked up in such a way that it can be solved explicitly all the way to the end. Given are the two points $A=(0,-7)$, $B=(-7,0)$, and the circle $\gamma: \>x^2+y^2=4$. Determine two points $P$, $Q\in\gamma$ such that the quantity $$d(P,Q):=|AP|^2+|PQ|^2+|QB|^2$$ becomes maximal, resp., minimal. You will ...


2

Here two other ones : Exercice 1 : Let $\alpha\in\mathbb{R}_{+}$. For all $\left(x,y,z\right)\in\mathbb{R}^{3}$, find the optimal constant $C\left(\alpha\right)\in\mathbb{R}$ such that $$x^{3}-\left(\frac{z}{\alpha}\right)^{3}\leq C\left(\alpha\right)\left(x^{2}+y^{2}+z^{2}\right)^{3/2}$$ and precise the case of equality in term of $\alpha$. ...


2

Sure, here's 2: a really good one to start out with is the optimization problem behind Principal Component Analysis. $$\text{max$_{\bf v}$ } \langle \bf{v},\Sigma_n \bf{v}\rangle$$ $$\text {s.t.}$$ $$||\bf {v}||_2=1$$ where $\bf v$ is a vector and $\Sigma_n= \frac 1n\sum_{i=1}^n(\bf x_i -\mu_n)(\bf x_i - \mu_n)^T$ is the Covariance matrix of the data ...


0

I was not familiar with ADMM and you sparked my interest. I found this paper under the main references at stanford university who have a nice homepage on the topic. Stated on page 13 of the paper you can see that the ADMM algorithm solves equality constrained problems of the form $$\min\ f(x)+g(z)$$ $$s.t. \ Ax+Bz=c$$ However, I will keep on reading to get ...


1

I think that it's the following. Define $f: M_m \to \mathbb{R}$ as $f(p)=|p-y|^2$, where $|\cdot |$ is the usual Euclidean norm on $\mathbb{R}^n$. Then $f$ is differentiable on $M_m$ because it is the restriction of a differentiable function on $\mathbb{R}^n$. Notice that for any $p \in M$ and $v \in T_pM$, the differential $df_p: T_pM \to \mathbb{R}$ is ...


0

Let's order the equations as 1, 2, 3, 4, 5. You probably get $x=y$ or $\mu=-0.5$ by plugging 3 into 4 and some manipulation. And that's correct. Now if $u=-0.5$, plug it into 3, 4, 5, you will get $$x+y-\lambda =0\\ 2+\lambda +z =0$$ Since 1 gives you $x+y=z$, you can get $\lambda =x+y=z $ and $\lambda=-z-2$. Combining these gives you $z=-1$. Then use ...


0

Here $\lambda$ is an unknown. If you prefer, you can switch to $\bar{\lambda}=-\lambda$ and get $F(t,u,\bar{\lambda})=h(t,u)-\bar{\lambda}g(t,u)$. If you find $\bar{\lambda}$, you find also $-\bar{\lambda}=\lambda$. From a geometric viewpoint, what you request is that the gradient of $h$ be parallel to the gradient of the constraint $g$: no matter if they ...


1

You have for $x>y$, $ a=0, b=\lambda$, and for $x<y$ $ a=\lambda , b=0$. Btw, your third derivative should be $c-h(x,y)=0$ and not $c-h_x =0$. Now write $c=h(x,y)$, for the first case, $c= y$, so the minimum value is $cb$, for the second case the minimum value is $ac$.


2

You have proved that $\lambda=1$ was impossible. So from (2): $$ 0 = (\lambda-1)y \implies y = 0; \\ x - 2 = \pm 1 $$ which gives the maximizer and minimizer.


3

Lagrange multipliers finds local extrema. On a domain with boundary, global extrema might be on the boundary instead. On an unbounded domain, the objective function might not be bounded above (resp. below), even on the surface where the constraint holds, in which case the maximum (resp. minimum) could be at infinity. In this case, assuming the problem ...


1

How about $(x,y,z)=(\cos t,\sin t,-\sin t\cos t)$ Check the derivative is never the zero vector.



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