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let $A\in <!--\; \in \; -->{\mathbb{R}}^{2x2}$ be symmetric matrix and C$\in <!--\; \in \; -->{\mathbb{R}}^{\mathbb{n}}$ a closed, convex cone whose linear hull is the entire ${\mathbb{R}}^2$ and
$x^{T}Ax>0\mathrm{\forall <!--\; \forall \; -->}x\in <!--\; \in \; -->C\mathrm{\setminus <!--\; \setminus \; -->}\{0\}$
but $A$ not positive semi-definite.

I constructed $A$ as follows:
$A=\left(\begin{array}{rr}1& 0\\ 0& 1\end{array}\right)$, such that $x^{T}Ax=x_1^2+x_2^2>0$
I am completely stuck on constructing a cone that will fullfill above conditions. Any input is highly appretiated.

Consider bellow LP constraints:
$$\begin{gathered} -<!--\; -\; -->1\ge <!--\; \ge \; -->x_1-<!--\; -\; -->x_4 \\ -<!--\; -\; -->4\ge <!--\; \ge \; -->x_2-<!--\; -\; -->x_1 \\ -<!--\; -\; -->9=x_2-<!--\; -\; -->x_3 \\ 5\ge <!--\; \ge \; -->x_3-<!--\; -\; -->x_1 \\ -<!--\; -\; -->3\ge <!--\; \ge \; -->x_4-<!--\; -\; -->x_3 \end{gathered}$$
How is it possible to show that the above constraints have no solution?

Find $x_1,x_2$ such that $\underset{x_1,x_2}{min}{x}_1^2+2x_1{x}_2$ where $x_1,x_2$ are subject to constraint $x_1^2{x}_2\ge <!--\; \ge \; -->10$.
I have changed the constraint into the equality $x_1^2{x}_2-<!--\; -\; -->10-<!--\; -\; -->s^2=0$ and attained the gradients which result in 4 equations and 4 unknowns:
$\begin{array}{rl}{x}_1^2{x}_2-<!--\; -\; -->10-<!--\; -\; -->s^2& =0\\ 2x_1+2x_2& =\mathrm{\lambda <!--\; \lambda \; -->}(2x_1{x}_2)\\ 2x_1& =\mathrm{\lambda <!--\; \lambda \; -->}{x}_1^2\\ 0& =\mathrm{\lambda <!--\; \lambda \; -->}(-<!--\; -\; -->2s)\end{array}$
But I am unsure of how to proceed from here. Additionally, I am struggling to find the dual problem.

I am finding the extreme points of the function $z=x^3+y^3-<!--\; -\; -->(x+y{)}^2$. This function has two critical points (0;0) and (4/3;4/3). The second point is an extreme point. For the first point, I have no conclusion. I claimed it is not an extreme point. Clear, $z<0$ when $x,y<0$. I am looking for $(x;y)$ such that $z>0$. Am I correct?

Suppose I have a problem of the form
$mi{n}_{x\ge <!--\; \ge \; -->\mathrm{ϵ<!--\; ϵ\; -->},y\ge <!--\; \ge \; -->0}\frac{f(x)+g(y)}{x+y}$
subject to some (convex) inequality constraints and some affine equality constraints, and where $f$ and $g$ are known to be convex, and $\mathrm{ϵ<!--\; ϵ\; -->}>0$ is some known constant. We can also assume that the feasible set is compact (in addition to being convex).

The main challenge here is when $\frac{f(x)+g(y)}{x+y}$ is not convex (otherwise it is just a convex problem), and also we don't have an easy escape like $f$ and $g$ are log convex, for example.

In my specific situation, I also know that $f,g$ are of the form
$f(x)={\int <!--\; \int \; -->}_0^x{h}_f(u)du,$
and
$g(x)={\int <!--\; \int \; -->}_0^x{h}_g(u)du,$
for some (known) nondecreasing and integrable (but not necessarily continuous) functions $h_f:[0,B_f]\to <!--\; \to \; -->\mathbb{R}$ and $h_g:[0,B_g]\to <!--\; \to \; -->\mathbb{R}$ for some real numbers $0<B_f,B_g<\mathrm{\infty <!--\; \infty \; -->}$.

Are there any easy ways to tackle this sort of problem? I am hoping to find ways that would be computationally not much more difficult than convex optimization, but I am not sure if this is possible... it seems like even the simpler problem of minimizing $f(x)/x$ is (surprisingly) difficult...

for the function f(x,y)= $\underset{i=1}{\overset{n}{\sum <!--\; \sum \; -->}}[(x_i{)}^{4\ast <!--\; \ast \; -->y_i}+(1-<!--\; -\; -->y_i)ln{x}_i]$, y are binary variables, x nonnegative real values with fixed upper bound C. Does the minimization min f(x,y) have a unique solution or not?

This is taken from one of the lecture slides, where a simplification of the quadratic formula is attempted:
$(x-<!--\; -\; -->x_A{)}^{T}A(x-<!--\; -\; -->x_A)+a^T(x-<!--\; -\; -->x_a)+b$
and from there I expand the brackets, which yields:
$(x-<!--\; -\; -->x_A{)}^{T}A(x-<!--\; -\; -->x_A)+a^T(x-<!--\; -\; -->x_a)+b=x^{T}Ax-<!--\; -\; -->x_A^{T}Ax-<!--\; -\; -->x_A^{T}A{x}_A+a^{T}x-<!--\; -\; -->a^T{x}_a+b$
In the next step the factorization is used to rearrange the formula:
$(x-<!--\; -\; -->x_A{)}^{T}A(x-<!--\; -\; -->x_A)+a^T(x-<!--\; -\; -->x_a)+b=x^{T}Ax+(-<!--\; -\; -->A^T{x}_A-<!--\; -\; -->x_A^T{A}^T+a{)}^{T}x+(x_A^{T}A{x}_A+b-<!--\; -\; -->a^T{x}_a)$
I do not understand how the linear term has been transposed from $(x_A^T{A}^T{)}^{T}x$. Can someone explain this ?

Most statements of Constraint Qualification I have found in the literature mention a locally "locally optimal solution" of the problem:
$\{\begin{array}{l}minf(x)\\ \text{s.t.}\\ g_i(x)\le <!--\; \le \; -->0\end{array}$
It is stated that when a C.Q. holds at a local optimum, then there exist Lagrange multipliers that satisfy KKT conditions.
But, I cannot get my head around this notion of local optimality. Does it mean locally optimal for the unconstrained problem? Does not local optimality imply the satisfaction of the KKT conditions?

Could you help me solve this problem please ?

1. Maximize $x^{t}y$ with constraint $x^{t}Qx\le <!--\; \le \; -->1$ (where $Q$ is definite positive)
What I tried : I tried using KKT but I don't know why I get $-<!--\; -\; -->\sqrt{y^t{Q}^{-<!--\; -\; -->1}y}$ as the maximum instead of $\sqrt{y^t{Q}^{-<!--\; -\; -->1}y}$ (which I believe is the maximum). Also, since $x^{t}y$ is linear (convex and concave), I don't know how to conclude...
2. Conclude that $(x^{t}y{)}^2\le <!--\; \le \; -->(x^{t}Qx)(y^t{Q}^{-<!--\; -\; -->1}y)$$\mathrm{\forall <!--\; \forall \; -->}x,y$ (generalized CS)

I want to ask whether the following question has a closed form solution:
$\{\begin{array}{l}{\displaystyle \underset{x}{min}{c}^{T}x}\\ \text{ s.t. }\\ {\mathbf{1}}^{T}x=b\\ x\ge <!--\; \ge \; -->0\end{array}$
We can write down the kkt condition as:
$\begin{array}{rl}L& =c^{T}x-<!--\; -\; -->⟨<!--\; ⟨\; -->{\mathrm{\lambda <!--\; \lambda \; -->}}_1,{\mathbf{1}}^{T}x-<!--\; -\; -->b⟩<!--\; ⟩\; -->-<!--\; -\; -->⟨<!--\; ⟨\; -->{\mathrm{\lambda <!--\; \lambda \; -->}}_2,x⟩<!--\; ⟩\; -->\\ \mathrm{\partial <!--\; \partial \; -->}L/\mathrm{\partial <!--\; \partial \; -->}x& =c-<!--\; -\; -->\mathbf{1}{\mathrm{\lambda <!--\; \lambda \; -->}}_1-<!--\; -\; -->{\mathrm{\lambda <!--\; \lambda \; -->}}_2=0\\ [{\mathrm{\lambda <!--\; \lambda \; -->}}_2{]}_i\cdot <!--\; \cdot \; -->x_i& =0\\ [{\mathrm{\lambda <!--\; \lambda \; -->}}_2{]}_i& \ge <!--\; \ge \; -->0\end{array}$
Denote ith component of $c$ as $c_i$. Now we can see that when $(c_i-<!--\; -\; -->{\mathrm{\lambda <!--\; \lambda \; -->}}_1)\ge <!--\; \ge \; -->0$, then we must have $x_i=0$ and $(c_i-<!--\; -\; -->{\mathrm{\lambda <!--\; \lambda \; -->}}_1)=0$, we also must have $x_i>0$. But I am not sure how to proceed to the next.