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Let's review this in $\mathbb R^3$: Consider any two points $(x,y,z)$ and $(a,b,c)$ in $\mathbb R^3$. The distance between them is $D=\sqrt { (x-a)^{2}+(y-b)^{2}+(z-c)^{2}}$. Now suppose $(a,b,c)$ is fixed and that $(x,y,z)$ lie in the plane $Ax+By+Cz=E$. Your problem is to find the point(s) on the plane such that $D$ is minimum. One way to do it is ...

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Your set compact set $K$ is stratified: It is a $3$-simplex consisting of an open interior, $4$ relatively open two-dimensional facets, $6$ relatively open edges, and $4$ vertices. In order to find the extrema of $h$ on $K$ we have to look at the $1+4+6+4=15$ faces in turn, in order to set up a "candidate list". As $$\nabla ... 0 Yes, you can use the Euler-Lagrange equations on the functional that you have written. You can then look for a pair of functions (y,λ(x)) such that y solves the Euler-Lagrange equations for the given λ(x) and at each point x one of the following is true: −y′(x)<0 and λ(x)=0 or −y′(x)=0 and λ(x)\geq0 Here is some intuition: If λ(x) is positive then this ... 0 I would diagonalize the covariance matrix \sum_i v_i^Tv_i first. That matrix is symmetric (and positive semidefinite), so it has an orthogonal eigenbasis. The eigendirections can be found by some standard numeric methods. After the orthonormal basis transform, the matrix \sum_i v_i^Tv_i is diagonal with consisting of the eigenvalues along the diagonal; ... 2$$f = xy^2$$with$$4=x^2+4y^2$$becomes$$f = xy^2 = x\left(1-\frac{1}{4}x^2\right) = x - 0.25x^3$$maximum via derivative$$\frac{d}{dx}f = 0 = 1 - 0.75x^2x_{1,2}=\pm\sqrt{\frac{4}{3}}\frac{d^2}{dx^2}f(x_1) < 0 \frac{d^2}{dx^2}f(x_2) > 0 $$and with 4=x^2+4y^2 again:$$y_{1,2} = \pm\sqrt{\frac{2}{3}}$$I think that's ... 2 You can suppose x,y>0 otherwise xy^2\leq 0. Then you can use that \frac{x_1+x_2+x_3}{3}\geq \sqrt[3]{x_1x_2x_3} and the equality holds iff x_1=x_2=x_3. Note that (x^2+2y^2+2y^2)^3\geq 27x^2y^4. So 64\geq 27(xy^2)^2 the eqaulity holds iff x=\sqrt{2}y or x=-\sqrt{2}y. Therefore xy^2\leq \frac{8}{3\sqrt{3}}. 1 HINT: consider the function$$L(x,y,z,\alpha,\beta)=xy+yz+\alpha(x+2y-6)+\beta(x-3z)$$we get by differentiating$$f_x=y+\alpha+\beta=0f_y=x+z+2\alpha=0f_z=y-3\beta=0f_{\alpha}=x+2y-6=0f_{\beta}=x-3z=0$$and we get$$\alpha=-2,\beta=\frac{1}{2},x=3,y=\frac{3}{2},z=1$$2 you must solve the system$$2x+\lambda(1+2x)=02y+\lambda(1+2y)=0x+y+x^2+y^2=12$$easy you will get$$x=y=-3,\lambda=-\frac{6}{5}$$or$$x=y=2,\lambda=-\frac{4}{5}$$1 Hint Highest (lowest) point is when z is maximal (minimal). Since you are on the paraboloid, z is fixed as a function of x,y. Plug this back into your constraint to get to optimize f(x,y) subject to g(x,y) = 12. 2 If you realize that the distance between the lines x+y=k and x+y=2 is just \frac{|k-2|}{\sqrt{2}} the problem boils down to computing the minimum and maximum of x+y over x^2+2y^2=1. Lagrange's method gives x=2y, hence (x,y)=\pm\frac{1}{\sqrt{6}}(2,1) and k=\pm\sqrt{\frac{3}{2}}. 1 If f(x,y) = x + y then \nabla f (x,y) = (1,1) orthogonal to the line x + y = 2. Considering \varphi (x,y) = x^2 + 2y^2 we have that \nabla \varphi (x,y) = (2x , 4x). Notice that 1 is a regular point of \varphi and M = \varphi^{-1} (1) is an ellipse, therefore compact. The restriction f|_M has then at least two critical points (Weiestrass ... 2 Write y(t)=\cos\big(x(t)\big) and z(t)=\sin\big(x(t)\big). Therefore, we may take x(0)=0 and x(1)=\left(2n+\frac12\right)\pi, for some integer n. The integral becomes \int_0^1\,\sqrt{1+\big(x'(t)\big)^2}\,\text{d}t, which calculates the length of the curve \big(t,x(t)\big) from t=0 to t=1. The critical curves are therefore straight ... 1 Another possibility. First,$$\tag{$*$} xy^2z^4 = 32 \Longrightarrow y^2 = \frac{32}{xz^4}. $$Notice that we can arbitrarily divide by x,y, or z since any point with a coordinate equal to 0 do not belong to your surface. Now, consider the squared norm of a generic point (x,y,z) and use (*):$$ x^2+y^2+z^2 = x^2 + \frac{32}{xz^4} + z^2 =: N(x,z). ...

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Your solution is correct, except the last part. You should get four minimizing points. (You mistakenly assumed that $y$ and $z$ must be both positive or both negative.) However, there is a solution without using Lagrange multipliers. Note by AM-GM that $$x^2+y^2+z^2=x^2+2\left(\frac{y^2}{2}\right)+4\left(\frac{z^2}{4}\right)\geq ... 0 A hint: Multiply (1) by x, (2) by y, and (3) by z and look at the three equations you got. 1 For (a), 1 is a regular value of f:\mathbb{R}^3\to\mathbb{R} defined by \left(x_1,x_2,x_3\right)\mapsto x_2x_3+x_3x_1+x_1x_2. That is, at every point p in f^{-1}(1), f is a submersion at p. In other words, \text{d}_pf:\text{T}_p\mathbb{R}^3\to\text{T}_1\mathbb{R} is surjective. This can be easily seen since the matrix representing ... 1 The simple solution (without KKT): it is easy to see that the problem is actually decoupled, i.e. the condition 0<x_k\le q affects only one term \log_2x_k in the objective function. It makes it possible to minimize over x_k each term independently, i.e.$$ \min\sum_{k=1}^n(-\log_2 x_k)=\sum_{k=1}^n\min_{0<x_k\le q}(-\log_2x_k)= ...

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Here's another approach for fun. We can use the AM-GM mean: $$\sqrt{ab}\leq \frac{a+b}{2},\quad \text{for all}\ a,b\geq 0$$ where equality holds if and only if $a=b$. Use $a=x^2,\ b=y^2$ to get: $$\sqrt{x^2y^2}\leq \frac{x^2+y^2}{2},$$ and our minimum will be where $x=y$. This implies $x=y=\pm\sqrt{3}$ and thus: $$2\sqrt{(xy)^2} = 2|xy| = 6\leq x^2+y^2,$$ ...

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There is also a geometric solution: the graph of $xy=3$ is a rectangular hyperbola, so we'll have to find the closest points of the hyperbola from the origin.

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HINT: Use $$\begin{cases} f_x = \lambda g_x \\ f_y = \lambda g_y \\ xy = 3\end{cases}$$ Note that $f_x = 2x, f_y = 2y, g_x = y, g_y = x$.

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Let $g(x,y)= x^2+y^2-25$. The Lagrange multiplier and required points can in other words be evaluated by: $$\lambda = \dfrac{f_x}{g_x}=\dfrac{f_y}{g_y}$$ Second order derivatives ( 6 in total) can be used to further establish min/max as with the single independent variable.

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Your extremal points are correct, but you made rather more work for yourself than you needed to. The Lagrange-multiplier method does tell us about the extremal points on the constraint circle. So we find a minimum value $\ -75 \$ at $\ (-3, \ 4) \$ and a maximum of $\ 125 \$ at $\ (3, \ -4) \$ . You are also correct in saying that we are not ...

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There was nothing "wrong" with your use of the Lagrange multipliers, but the method often leaves us with a set of equations that need to be "handled the right way" in order to be fully useful. (There is no general method for finding this "right way", since there are many ways that the system of equations can present us with an algebraic tangle. I am also ...

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The method of Lagrange's multipliers is a theorem with a few assumptions. Check it out on your favorite advanced calculus book. But, what is even more important, this method works for constrained critical points and should not be seen as an alternative to solving $\nabla f=0$.

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I'm not entirely sure what you're getting at, but I think this maybe helpful to you: The intuition is that an extreme point of $f(x)$ under the condition $g(x) = 0$ must satisfy $g(x) = 0$ and $\nabla_\nu f(x) = 0$ for any direction $\nu$ tangential to the candidate set $\{g(x) = 0\}$. If this were not the case, we could go a small, positive distance along ...

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The function can be written as $$z=\pm\sqrt{xy+1}.$$ Obviously, if $xy<-1$ then there is no real $z$. Note that $xy<0$ if the signs are different. This means that our function has limitations only on the second and the fourth quadrant. For a positive $x$ we have to find those negative $y$'s for which $xy<-1$. So, we have no solution if ...

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$A = x^{4} + y^{4} - 4b^{2}xy + 2b^4$ $A = (x^{2} - b^{2})^2 + (y^{2} - b^{2})^2 + 2x^{2}b^2 + 2y^2b^2 - 4b^{2}xy$ $A = (x^{2} - b^{2})^2 + (y^{2} - b^{2})^2 + 2b^2(x^2 + y^2 -2xy)$ $A = (x^{2} - b^{2})^2 + (y^{2} - b^{2})^2 + 2b^2(x - y)^2$ Which is positive as a sum of squares

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Hints: let $f(x,y) = x^4+y^4-4b^2xy$. Show that the critical points (gradient equals zero) of $f$ are $(b,b)$ and $(-b,-b)$, show that they are local minima, and note that $f$ at these points takes the value $-2b^4$.

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Let the cost of 1 square metre of plywood be 1 (the actual amount is irrelevant, we just need to set a base unit). So the cost of 1 square metre of glass is 2. Now let the dimensions of the box be $x$, $y$, and $z$ metres, with the glass side being one of the two $x \times y$ sides. Then the cost of the box is $C(x, y, z) = 3xy + 2xz + 2yz$ (because you ...

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I think there is nothing more to say in such a constraint optimisation problem. However, you can check whether the function actually does what it says it does. For instance, to show that they are maximum points, you convert the function to one-variable equation using the constraint, then differentiate once, and substitute. If the values are -ve, then you're ...

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