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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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@Michele_Laino Is this one A?
I'm sorry, here we have to apply the Kuhn-Tucker theorem, and I'm not good with that theorem
@dan815 please help

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As a start sketch the given inequalities : \(x\ge 1\) |dw:1443889526891:dw|
\(y\ge 0\) |dw:1443889681232:dw|
Ok so is the minimum 1?
\(x\le 4-y\) : |dw:1443889782148:dw|
As you can see, the intersection of all the given inequalities is : |dw:1443890083460:dw|
The max and min values of given function, \(f= 2y+x\) occur at the "vertices" of that intersection region. Look at that intersection region, could you find the vertices of that intersection region ?
No, how would I do that?
Easy, do you see intersection region in above diagram ?
Yes
|dw:1443890318233:dw|