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25 votes

Distance of ellipse to the origin

You can always try to write it in polar coordinates: $$3r^2+4r^2\sin\theta\cos\theta=20$$ $$r^2=\frac{20}{3+2\sin2\theta}\ge\frac{20}{3+2}=4$$
ajotatxe's user avatar
  • 65.2k
16 votes
Accepted

Distance of ellipse to the origin

Hint: $4xy \le 2(x^2+y^2), 3x^2+3y^2 = 3(x^2+y^2) \implies 5(x^2+y^2) \ge 20 \implies x^2+y^2 \ge .....$
DeepSea's user avatar
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10 votes
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Smallest value of $x^2+5y^2+8z^2$ given $xy+yz+xz=-1$.

Let $k$ be a minimal value. Thus, $k>0$ and $$x^2+5y^2+8z^2\geq k$$ or $$x^2+5y^2+8z^2+k(xy+xz+yz)\geq0$$ or $$x^2+k(y+z)x+5y^2+8z^2+kyz\geq0,$$ for which we need $$k^2(y+z)^2-4(5y^2+8z^2+kyz)\...
Michael Rozenberg's user avatar
9 votes
Accepted

Minimizing a quadratic function subject to quadratic constraints

Hint: Your conclusion that your three equations imply that there is no consistent choice for $\lambda$ only holds if $x$, $y$ and $z$ are non-zero! If any of them are zero, then their corresponding ...
Shinja's user avatar
  • 1,355
9 votes

What's the shortest distance from a point inside of an ellipsoid to its surface?

If the point is $p$ and the ellipsoid is $x^T Q x = 1$, you want to minimize $(x-p)^T(x-p)$ subject to $x^T Q x = 1$, where $Q$ is a (symmetric) positive definite matrix. Using a Lagrange multiplier, ...
Robert Israel's user avatar
7 votes

Distance of ellipse to the origin

This ellipse is centered at the orgin. The distance you seek equals the length of the minor axis. Next thing to know is that it has been rotate 45 degrees. $x = \frac {\sqrt 2}{2} x' + \frac {\sqrt ...
Doug M's user avatar
  • 58k
7 votes
Accepted

Find maximum of $C=2(x+y+z)-xy-yz-xz$

Let $(x+y+z) = a$ . Then $$(x+y+z)^2 = a^2 \implies (xy+yz+zx)=\dfrac{a^2-3}{2}$$ So $$2(x+y+z)-xy-yz-xz = 2a -\dfrac{a^2-3}2$$ $$ C = \dfrac{-a^2+4a+3}{2}$$ The maximum of this quadratic is at $a =...
The Demonix _ Hermit's user avatar
6 votes
Accepted

Maximization of quadratic form over unit Euclidean sphere not centered at the origin

This is (a variant of) the trust-region problem, and the solution can be computed rather easily, although not an analytical solution See, e.e., https://www8.cs.umu.se/kurser/5DA001/HT07/lectures/...
Johan Löfberg's user avatar
6 votes
Accepted

Find largest possible value of $x^2+y^2$ given that $x^2+y^2=2x-2y+2$

Hint You found $$(x-1)^2+(y+1)^2=4$$ which is the equation of a circle centered in $(1,-1)$ and with radius $2$. The function $x^2+y^2$ is the square of the distance of $(x,y)$ to the origin. Which ...
StackTD's user avatar
  • 27.9k
6 votes

Why does the Lagrange multiplier $\lambda$ change when the equality constraint is scaled?

If $x^\star$ minimizes $f(x)$ subject to the constraint that $g(x)=0$, then under mild assumptions there exists a Lagrange multiplier $\lambda$ that satisfies $$ \tag{1} \nabla f(x^\star) = \lambda \...
littleO's user avatar
  • 52.2k
6 votes

Minimizing $x^2+y^2+z^2$ subject to $xy -z + 1 = 0$ via Lagrange multipliers

$$x^2+y^2+z^2=x^2+y^2+z^2+2(xy-z+1)=(x+y)^2+(z-1)^2+1\geq1.$$ The equality occurs for $x=y=0$ and $z=1$, which says that we got a minimal value.
Michael Rozenberg's user avatar
5 votes

Distance of ellipse to the origin

By inspection (or see this for additional reference if required), we can see that this is an ellipse with centre at origin and axes of symmetry $y=\pm x$. Substituting into the ellipse equation gives ...
Hypergeometricx's user avatar
5 votes

Minimizing a quadratic function subject to quadratic constraints

When you solve the system with Lagrange method, your variables are $x,y,z$ and $\lambda$. One solution to the system is $$(x,y,z,\lambda)=(1,0,0,\frac{1}{4}),$$ which matches your intuitive solution. ...
Kuifje's user avatar
  • 9,594
5 votes

If $x$ and $y$ are real number such that $x^2+2xy-y^2=6$, then find the minimum value of $(x^2+y^2)^2$

Use $$2(x^2+y^2)^2-(x^2+2xy-y^2)^2=(x^2-2xy-y^2)^2\geq0.$$
Michael Rozenberg's user avatar
5 votes

Maximize $\mathrm{tr}(Q^TCQ)$ subject to $Q^TQ=I$

Since $C$ is symmetric real we can write $C=U \Lambda U^T$ where $\Lambda$ is a diagonal matrix of eigenvalues. As $Q^T U U^T Q = I$, we can just assume $C= \operatorname{diag} (\lambda_1,...,\...
copper.hat's user avatar
  • 173k
5 votes
Accepted

Box-constrained QCQP

This is really more of a comment than a complete answer, but in case you are not aware, a slightly simpler version of your quadratic program can be reformulated as, \begin{align} \text{minimize}\;\;&...
Set's user avatar
  • 7,630
4 votes

Can't find minimum using Lagrange multipliers

To find the extreme values, you need to also check the conditions where $f_x=0$ and $f_y=0$, which is the way to find local extremas before using Lagrange. In this case you can find through $f_y=0$, ...
Bridget's user avatar
  • 216
4 votes

Distance of ellipse to the origin

Taking into account that it is an ellipse with short axis $y=x$ find crossings of this line with the ellipse.
Widawensen's user avatar
  • 8,224
4 votes

A convex optimisation problem involving the Euclidean norm

You are computing the matrix $2$-norm of the matrix $\begin{bmatrix} A \\ B \\ C \end{bmatrix}$. So just use any appropriate method from numerical linear algebra for computing the norm of a matrix. ...
littleO's user avatar
  • 52.2k
4 votes
Accepted

Minimizing Quadratic Form with Norm and Positive Orthant Constraints

I would use the Projected Gradient Descend for this case. Though the problem isn't Convex it will work nicely. The algorithm is as following: Calculate the Gradient at the current point. Update the ...
Royi's user avatar
  • 8,799
4 votes

Box-constrained QCQP

The problem that you have can be written as: $$ \begin{array}{ll} \underset {{\bf x} \in \mathbb{R}^n} {\text{minimize}} & \langle {\bf A}, \, {\bf X}\rangle + \langle {\bf a}, {\bf x}\rangle \\ \...
Kroki's user avatar
  • 13.2k
3 votes

Distance of ellipse to the origin

we have $$d(x,y)=\sqrt{x^2+y^2}$$ we get that $y$ from the curve: $$y_{1,2}=-\frac{2}{3}x\pm\sqrt{\frac{20}{3}-\frac{5}{9}x^2}$$ can you proceed?
Dr. Sonnhard Graubner's user avatar
3 votes
Accepted

Can't find minimum using Lagrange multipliers

You correctly got the three equations for the Lagrange method. The 2nd equation tells you that $\lambda=1$ or $y=0$. $\lambda=1$ gives you the stationary point you identified, which has $x=\frac{1}{...
almagest's user avatar
  • 18.4k
3 votes

Showing that $x^{\top}Ax$ is maximized at $\max \lambda(A)$ for symmetric $A$

There is a cleaner proof altogether that circumvents the need to consider the square root of $A$. Consider the spectral decomposition of $A$ given by $A = V\Lambda V^{\top}$. For some $x \in \mathbb{R}...
Ryan's user avatar
  • 850
3 votes

Quadratic optimisation with quadratic equality constraints

You can have your own opinion as to whether this is efficient in some sense. But if you want to solve the problem, then finding a solution can be considered more efficient than not finding a solution. ...
Mark L. Stone's user avatar

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