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You might like to check out the envelope theorem. A summary is here: https://en.wikipedia.org/wiki/Envelope_theorem Say $f(x,t)$ has positive partial derivative with respect to $t$. Then, if your feasible set defined by $G(x,t)\geq 0$ is not shrinking as $t$ gets larger, i.e. $G_t(x,t)\geq 0$, then if the mild conditions that are described in the second ...


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Your dual function $$g(\lambda) := \min_{\lambda} \left[ 2-\lambda+( -x-\lambda)x\right] $$ is better written $$ h(\lambda) := \max_{x} \left[ 2-\lambda+x( \lambda - x)\right]\tag1 $$ because (a) you've used $g$ already for $g(x)=x-1$, and (b) you should be maximizing over $x$, not minimizing over $\lambda$, and (c) you have a sign wrong in your ...


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Completing squares I get, with $\;x,y,z\;$ instead of $\;x_i\;$ : $$3x^2-2xy+2y^2-2yz+3z^2=3\left(x-\frac13y\right)^2+\frac43y^2+3\left(z-\frac13y\right)^2\implies$$ $$\implies \begin{cases}I\;\;\;\;\,a=x-\frac13y\\II\;\;\;b=y\\III\;c=z-\frac13y\end{cases}\;\;\; \implies\;\;\begin{cases}x=a+\frac13b\\y=b\\z=c+\frac13b\end{cases}$$ So the base part is ...


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I'm not fully sure I understand the question. I see a few interpretations: Are stationary points of $f$ stationary points of $g$? The answer is very much no: for instance, $g(x,y)=xy$ only has a stationary point at $(0,0)$, but that's not the only stationary point we find when we maximize $g$ subject to $2x+2y=10$. However, stationary points of $g$ which ...


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In general, no. The method of Lagrange multipliers finds points on the level surface $h=\text{const}$ at which the gradients of $g$ and $h$ are parallel, i.e., where $\nabla f$ is normal to this surface, but that need not happen at a stationary point of the unrestricted function $g$. Indeed, $g$ may not have any stationary points at all. Consider, for ...


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The point-wise maximum of a set of convex functions is convex. $$g(x) = \max_i f_i(x) $$ The corresponding epigraph $$\{(x,t) | t > \max_i f_i(x)\}$$ is convex which could also be visualized as the insertion of a family of convex spaces, $$\cap_i \{(x ,t)| t > f_i(x)\}$$ The lagrangian could be rewritten as the negative of the supremum, ...


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We need the $\lambda_i$s positive in order to penalize a violation of the constraints $f_i(x)\leq 0$. More precisely, if they were negative, minimizing $L$ would likely give $x$s for which $f_i(x)>0$ (because then $\lambda_if_i(x)<0$ making $L$ smaller) which is in violation to our primal problem. The final inequality holds by construction of the $L$. ...


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I have dug up some book titles that i found useful, so i am writing my own answer. Namely: -Mathematical methods for physics and engineering by Riley, Hobson, Bence. -Many books or chapters from books of the author Vladimir Arnold. -Strang's Calculus -Hildebrand's Methods of Applied Mathematics -Churchill's Fourier Series and Boundary Value Problems The ...


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I recommend checking out Gilbert Strang's book Introduction to Applied Math. Strang has great intuition and I think he explains things in a very clear, simple, and yet deep way.


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Yes, it can still change. I guess that, in general, it is more likely that it will change than it not changing.


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For anyone interested, I was able to come to a seemingly valid solution using Evgeny's help. The reference I used is here. If you find any mistakes in my process, please point it out or write a correct answer, and I'll accept it instead of my own. Using the Bordered Hessian matrix: $H = \left[\begin{array}{c}0&g_x&g_y\\g_x&f_{xx}+\lambda ...


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This is a strange problem. Why are we not allowed to use optimization methods to solve optimization problem? But anyway ... First, use the geometric result that the area of a rectangle is maximized subject to perimeter constraint is a square (this itself need a proof using calculus), but look like we are allowed to use calculus so we will take this for ...


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If all constraints are linear, redundancy does not invalidate the validity of Lagrange multipliers (KKT conditions) for an optimum, but may result in non-uniqueness of the Lagrange multipliers, which could affect their interpretation as dual (constraint) prices. If one or more constraints are nonlinear, and the constraint gradients are not linearly ...



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