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Hot answers tagged lagrange-multiplier

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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 ...

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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.

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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 ...

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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 ...

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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 ...

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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 $\alpha$. ...

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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 ...

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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 ...

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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 ...

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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 ...

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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 ...

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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. ...

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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 $f:\mathbb ... 1 I don't think there is a very simple method to get the answer. But you can simplify the cases: case 1 :$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}), ...

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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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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 $ac$.

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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 ...

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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 ...

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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 $$\frac{\partial f}{\partial a}=\frac n2a^{n-1}-\frac{\lambda n}{2^n}(a+b)^{n-1}=0,\frac{\partial ... 1 Hint : Calculate the maximum of (\frac{x+y}{2})^n and the minimum of \frac{x^n+y^n}{2} with Lagrange multipliers and compare the results. 1 (I know this question is ancient, but I happened to run into it while looking for something else.) While I am not sure if S_{n,k} is concave on the probability simplex, you can prove the result you want and many other similar useful things using Schur concavity. A sketch follows. A vector y\in \mathbb{R}_+^n majorizes x \in \mathbb{R}_+^n if the ... 1 I assume you mean the constraint x^T Px + 2c^Tx + s \leq 0. For intuition on the difficulty of this constraint, let us assume we also have constraints x_i \in [0,1] for all i \in\{1, \ldots, n\}. Now consider your single constraint in the special case P=-I, c=(1/2, \ldots, 1/2), s=0:$$ -x^Tx + 1^Tx \leq 0  This is equivalent to saying: ...

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