Silly Quadratic programming

Question

Suppose that I want to minimize the function $x^2$ subject to the contraint $$ ax\leq b, $$ for some $b> 0$. I solved the problem if the contraint is an equality but I'm not sure how to go about solving the slack case... I looked at KKT but I'm confused....

Answer 1

First, suppose $b \ge 0$.

Then $x=0$ satisfies the contraint $ax\le b$, regardless of the value of $a$.

Hence, even with the constraint $ax\le b$, the minimum value of $x^2$ is $0$.

Next, suppose $b < 0$.

Consider cases . . .

If $a = 0$, the inequality $ax\le b$ reduces to $0\le b$, which is impossible (since $b < 0$), hence the feasible region is empty.

If $a > 0$, the inequality $ax\le b$ reduces to $x\le {\large{\frac{b}{a}}}$, hence, since ${\large{\frac{b}{a}}} < 0$, the minimum value of $x^2$ is $\bigl({\large{\frac{b}{a}}}\bigr)^2$.

If $a < 0$, the inequality $ax\le b$ reduces to $x\ge {\large{\frac{b}{a}}}$, hence, since ${\large{\frac{b}{a}}} > 0$, the minimum value of $x^2$ is $\bigl({\large{\frac{b}{a}}}\bigr)^2$.