# Solving inequality using matrix form condition

*answer is $$x=(10,5)^T$$
(a,b) is inner product.
x and y is 2d vector.

Find $$\bar{x}$$ within the area by the systems of inequality,$$K(x)$$ s.t.

About inner product of vector F(x) and (y-x),
$$F(\bar{x})^T.(y-\bar{x}) \geq 0$$ For Any $$y$$ in $$K(\bar{x})$$

$$F(x)=a.x-b$$
$$K(x)=\bar{K}\space and \space \hat{K}$$
$$\bar{K}=c.y-d\leq 0$$
$$\hat{K}=e.y+g.x-h\leq 0$$

first,setting the matrix and vector

a = {{2, 8/3}, {5/4, 2}};
b = {{34}, {24.25}};
c = {{-1, 0}, {1, 0}, {0, -1}, {0, 1}};
d = {{0}, {11}, {0}, {11}};
e = {{1, 0}, {0, 1}};
g = {{0, 1}, {1, 0}};
h = {{15}, {15}};


so I want to do like the following)

y = Array[yy, {2}];
x = Array[xx, {2}];
cond1 = # <= 0 & /@ Flatten@(c.y - d);
cond2 = # <= 0 & /@ Flatten@(e.y + g.x - h);
F = Transpose[a.x - b];
target = First@Flatten@(F.(y - x));
SomeFunction[target>=0, And @@ Join[cond1, cond2]]


=>

{xx[1]->10,xx[2]->5}


how to solve the inequality F.(y-x)>=0 using cond1 and cond2?
this returns error:

Solve[{target >= 0, List @@ Join[cond1, cond2]}, Join[x, y]]


Using your definitions of a, b, c, d, e, g, h, target, cond1, cond2, you can find the minimum value of target (under the constraints) using NMinimize:

NMinimize[{target^2, Join[{cond1, cond2}]}, Flatten[{x, y}]]
{3.29871*10^-22, {xx[1] -> 7.92824, xx[2] -> 6.96093,
yy[1] -> 0.530959, yy[2] -> -0.455972}}


I don't think this is exactly what you are looking for, but it might help to get you started.

Because this kind of dual problem should be treated as a multi-step decision problem, I use this as the answer to the question.

$$L(y,x):=F(x)^T.(y-x)$$
$$L(y,x)\geq 0 \Rightarrow minL(y,x)\geq0$$

Avoiding the case of minL==0, because of constraints.

Table[{s, j,
First@Flatten[
NMinimize[{(target /. {xx[1] -> s, xx[2] -> j}),
List @@ Reduce[
And @@ Join[cond1, cond2] /. {xx[1] -> s, xx[2] -> j}, y]},
y]]}, {s, 0, 15}, {j, 0, 15}] // Flatten[#, 1] & //
Select[#, Last@# > 0 &] & // First@SortBy[#, Last@# &] &


{10, 5, 1.77636*10^-15}