Question

Consider the function \[ f(x,y) = x^2y + 2xy^2 - 2x^2y^2. \] Then which one of the following statements is correct?

• A. $\left(\tfrac{3}{2},0\right)$ is a point of local maxima of $f$.
• B. $\left(0,\tfrac{3}{4}\right)$ is a point of local minima of $f$.
• C. $\left(\tfrac{1}{2},\tfrac{3}{4}\right)$ is a point of local maxima of $f$.
• D. $\left(\tfrac{3}{2},\tfrac{3}{4}\right)$ is a saddle point of $f$.

Hint:

To classify critical points in multivariable calculus, always use the Hessian determinant test: $D>0$ with $f_{xx}>0$ = local min, $f_{xx}<0$ = local max; $D<0$ = saddle point.

Answer 1

The Correct Option is D

Step 1: Compute first partial derivatives.
\[ f_x = 2xy + 2y^2 - 4xy^2, f_y = x^2 + 4xy - 4x^2y. \] 

Step 2: Find critical points.
Set $f_x=0, f_y=0$. Solve: At $(x,y)=(3/2,3/4)$ both vanish (verified). 

Step 3: Compute second derivatives.
\[ f_{xx} = 2y - 4y^2, f_{yy}=4x - 4x^2, f_{xy}=2x+4y-8xy. \] At $(3/2,3/4)$: \[ f_{xx}=2(0.75)-4(0.75)^2=1.5-2.25=-0.75, \] \[ f_{yy}=4(1.5)-4(2.25)=6-9=-3, \] \[ f_{xy}=2(1.5)+4(0.75)-8(1.125)=3+3-9= -3. \] 

Step 4: Hessian determinant.
\[ D = f_{xx}f_{yy} - (f_{xy})^2 = (-0.75)(-3) - (-3)^2 = 2.25-9=-6.75 <0. \] Negative determinant $\Rightarrow$ saddle point. Final Answer:
\[ \boxed{\left(\tfrac{3}{2},\tfrac{3}{4}\right)\ \text{is a saddle point.}} \]