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

Step 1: Find the critical points. \[f_x=3x^2-3y=0 \implies y=x^2,\qquad f_y=3y^2-3x=0\implies x=y^2.\] Substituting \(y=x^2\) into \(x=y^2\) gives \[x=(x^2)^2=x^4 \implies x^4-x=0\implies x(x^3-1)=0\implies x(x-1)(x^2+x+1)=0.\] The factor \(x^2+x+1\) has discriminant \(1-4=-3<0\), so it has no real roots. The only real solutions are \(x=0\) and \(x=1\), giving critical points \[(0,0)\quad\text{and}\quad(1,1).\] So f has exactly two critical points, not three.

Step 2: Classify each critical point with the second derivative test. \[f_{xx}=6x,\qquad f_{yy}=6y,\qquad f_{xy}=-3,\qquad D=f_{xx}f_{yy}-f_{xy}^2.\] At (0,0): \(f_{xx}=0,\ f_{yy}=0,\ D=0\cdot 0-(-3)^2=-9<0\), so (0,0) is a saddle point. At (1,1): \(f_{xx}=6,\ f_{yy}=6,\ D=36-9=27>0\) and \(f_{xx}>0\), so (1,1) is a local minimum.

Step 3: Compare with the statements. There is exactly one saddle point, at (0,0), and exactly one local minimum, at (1,1); there is no local maximum anywhere. Statement (I), which claims 3 critical points, is false, and so is the local-maximum claim in (III). The classification that matches the true picture, one saddle point together with one local minimum, corresponds to statement (II). \[\boxed{\text{Only (II) is true}}\]