Question 529703
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This question has no answer.  That is because your constraint inequalities are strictly less than rather than less than or equal.  That is because the optimum point, if it exists at all, lies on a vertex of the quadrilateral bounded by the boundary lines of the four constraint inequalities. But since you have made the constraints strictly less than, you can pick a point as close as you like to the vertex of your feasiblilty area, and  then I can find one closer, after which you can find one even closer, ad infinitum.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
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