Graduate Finding dual optimum of a linear problem

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The discussion focuses on finding the dual optimum of a linear problem defined by affine constraints. The Lagrangian function is derived as c'x + λ'(Dx-e) when constraints are transformed. By differentiating the Lagrangian with respect to x, it is established that c = -D'λ. The dual function g(λ) is then expressed as -λ.e, leading to the conclusion that d* equals the supremum of g(λ). The query raised is whether this differentiation method yields the dual optimum for all convex inequality constraints and convex objective functions.
Trollfaz
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A simple linear problem goes
min c'x such that f_i(x)<= 0 and Ax=b
x
Suppose we make all constraints affine. Then

Dx-e<=0 and Ax-b =0
We get the Langrangian function as
c'x + λ'(Dx-e) +ν'(Ax-b) and since Ax-b is 0,
we reduce L to
c'x + λ'(Dx-e)
The dual function g is
inf L(x,λ)
x
Then I differentiate L against x to get c=-D'λ
With that we get g(λ) as -λ.e
So I conclude that d*=sup g(λ) against λ.
Does differentiation of the Lagrangian function give me the dual optimum and does it work for all convex inequality constraints and convex objective functions?
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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