Minimize (xy-1)^2 + x^2

Question

I am practicing for my final exam and I came across this exercise online. Yet, I have no clue how to solve it.

min (xy-1)^2 + x^2

I figured the four step method would be a good idea. I got stuck on how to prove Weierstrass. Also, the solution I got with the First Order Necessary Conditions is the answer 1. From the plot, however, I can see there should be multiple solutions to this exercise. Can anyone help me?

score 2 · Accepted Answer · answered Dec 12 '17 at 13:40

2

Let $f(x,y)=(xy-1)^2+x^2$. Then $f(x,y) \ge 0$ for all $(x,y)$.

For $x \ne 0$: $f(x, \frac{1}{x})=x^2 \to 0$ as $x \to 0$.

Hence $ \inf \{f(x,y):(x,y) \in \mathbb R^2\}=0$, but there is no $(a,b)$ such that $f(a,b)=0$.

Conclusion: $ \min \{f(x,y):(x,y) \in \mathbb R^2\}$ does not exist !

answered Dec 12 '17 at 13:40

Fred

77,394

Thank you very much for your answer! But how can I prove that the function is not coercive? – Anonymous Dec 12 '17 at 13:42

Minimize (xy-1)^2 + x^2

1 Answers1