Minimize $x^2+xy-x+y$ subject to $x,y > 0$

Question

$$\begin{array}{ll} \text{minimize} & x^2+xy-x+y\\ \text{subject to} & x, y \geq 0\end{array}$$

Gradient cannot be zero since $x$ and $y$ are non-negative.I know the minimum value is -1/4 and I do not want the minimum using inequalities.I am wondering how I can show at the minimum point there is no direction for improvement.Question from Jalil Hajimir

Answer 1

Write your term in the form $$\left(x-\frac{1}{2}\right)^2+y(x+1)-\frac{1}{4}$$ so the searched minimum is given by $-\frac{1}{4}$ and will be reached for $$x=\frac{1}{2},y=0$$

Answer 2

Another solution:

It is clear that the minimun will not be obtained for large values of $x$ of $y$.

Let us bound the set: e.g. $x \le X$ and $y \le Y$.

The new set is closed and bounded.

As $f(x, y)$ is continue and derivable, at least one global minimum exits in this set, it must correspond to a local minimum (null gradient) or must be on the boarder.

As the gradient cannot be nulled in the set, then the absolute minimum must correspond to $x = 0$ or $y = 0$.

It $y = 0$, the function becomes $x^2 - x$, the minimum of which is $-\frac{1}{4}$ for $x = \frac{1}{2}$.

For $x = 0$, we get $f(x = 0, y) = y$, the minimum of which is $0$, which is higher than $- \frac{1}{4}$

Answer 3

You can use the KKT conditions... But you can also note that since the objective function is differentiable the extrema will occur either on the boundary of the feasible set ($x=0$ or $y=0$) or at critical points in the interior of the feasible. As you noted, the second situation cannot occur. If $x=0$, the minimum value of $f$ is attained with $y=0$ and is zero. If $y=0$, the values of $f$ are given by $x^2-x$, that attains a global minimum for $x= \frac 12$.