3
$\begingroup$

I would like to find the maximum of the function $f(x,y)=x^2+y^2$ on the constraint $x^2-y^2=(x^2+y^2)^2$. The level curve $h(x,y)=0$ of the function $h(x,y)=x^2-y^2-(x^2+y^2)^2$ looks like an infinity sign and, because of the radial symmetry of $f$, it is clear that it takes its maximum value at (-1,0) and (1,0).

My question is: How can I show this using Lagrange multipliers?

enter image description here

$\endgroup$

2 Answers 2

1
$\begingroup$

You first define the Lagrangian $L(x,y, \lambda) = x^2+y^2 + \lambda (x^2-y^2 - (x^2+y^2)^2).$ Then, the KKT conditions are given by: $$\begin{cases} \nabla f(x^{\star}, y^{\star}) + \lambda \nabla h(x^{\star},y^{\star}) = 0\\ h(x^{\star},y^{\star}) = 0\end{cases}.$$ If you compute the optimality conditions and primal feasibility you get the following system of equations:$$\begin{cases} 2x+2\lambda x - 4\lambda x^{3} - 4\lambda xy^{2} = 0\\ 2y-2\lambda y - 4\lambda y^{3} - 4\lambda yx^{2} = 0\\ x^{2} - y^{2} - x^{4} - y^{4} - 2x^{2}y^{2} = 0 \end{cases}$$ which gives you the solutions you intuitively found.

$\endgroup$
0
$\begingroup$

An alternative way without using Lagrange multipliers: notice that if $x^2-y^2=(x^2+y^2)^2$ it is impossible that $x^2+y^2>1$, as we would have $$(x^2+y^2)^2> x^2+y^2\geq x^2-y^2$$

Thus, we will have at most $x^2+y^2=1$. But this equality is reached at $x=1,y=0$ and $x=-1,y=0$, so the maximum of $x^2+y^2$ at the given curve is $1$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .