Question

For the function \(f(x,y)=x^3+y^3-3xy\), consider the following statements:
(I) f has 3 critical points.
(II) f has two saddle points and one point for local minimum.
(III) f has only one saddle point, a point for local minimum and a point for local maximum.
Choose the correct answer.

• A. Only (I) is true
• B. Only (II) is true
• C. Only (I) and (II) are true
• D. Only (I) and (III) are true

Hint:

Solve f_x=0 and f_y=0 simultaneously to locate all real critical points, then classify each with the Hessian test.

Answer 1

The Correct Option is B

An alternate way to locate the critical points. From $f_x=0$ and $f_y=0$: \[x^2=y,\qquad y^2=x.\] Subtracting, $x^2-y^2=y-x$, so $(x-y)(x+y)=-(x-y)$, which gives \[(x-y)(x+y+1)=0.\] So either $x=y$, or $x+y=-1$.

Case 1: $x=y$. Then $x^2=y=x$, so $x^2-x=0$, $x(x-1)=0$, giving $x=0$ or $x=1$. This produces the points $(0,0)$ and $(1,1)$.

Case 2: $x+y=-1$, i.e. $y=-1-x$. Substituting into $x^2=y$: $x^2=-1-x$, so $x^2+x+1=0$, whose discriminant is $1-4=-3<0$. No real solution here.

So $f$ has exactly two real critical points, $(0,0)$ and $(1,1)$.

Classify using Hessian eigenvalues. The Hessian is $H=\begin{pmatrix}6x & -3\\ -3 & 6y\end{pmatrix}$.
At $(0,0)$: $H=\begin{pmatrix}0&-3\\-3&0\end{pmatrix}$, eigenvalues solve $\lambda^2-9=0$, so $\lambda=\pm3$. Opposite signs mean $H$ is indefinite, so $(0,0)$ is a saddle point.
At $(1,1)$: $H=\begin{pmatrix}6&-3\\-3&6\end{pmatrix}$, trace $=12$, determinant $=27$, so $\lambda^2-12\lambda+27=0$ gives $\lambda=9$ or $\lambda=3$, both positive. $H$ is positive definite, so $(1,1)$ is a local minimum.

This confirms one saddle point and one local minimum, with no third critical point and no local maximum, matching statement (II). \[\boxed{\text{Only (II) is true}}\]