Question

Consider the function \(f:\mathbb{R}^2 \to \mathbb{R}\) given by

• A. \(f\) has exactly \(4\) critical points
• B. \(f\) has more than one saddle point
• C. \(f\) has two local minima
• D. \(f\) has a critical point of the form \((a,b)\) such that \(a+b=\frac{16}{15}\)

Hint:

For a function of two variables, first find critical points using \(f_x=0\) and \(f_y=0\), then classify them using the Hessian determinant \(D=f_{xx}f_{yy}-(f_{xy})^2\).

Answer 1

The Correct Option is C

Step 1: Find the partial derivatives.
With $f=xy(5x+3y-6)$, $f_x=y(10x+3y-6)$ and $f_y=x(5x+6y-6)$.

Step 2: Solve for critical points.
Setting both to zero gives $(0,0)$, $\left(\tfrac65,0\right)$, $(0,2)$ and $\left(\tfrac25,\tfrac23\right)$, so there are exactly $4$, and (A) holds.

Step 3: Classify with the Hessian.
$f_{xx}=10y$, $f_{yy}=6x$, $f_{xy}=10x+6y-6$. At the three points with a zero coordinate, $D=f_{xx}f_{yy}-f_{xy}^2=-36<0$, so all three are saddles. That confirms (B).

Step 4: Check the inside point.
At $\left(\tfrac25,\tfrac23\right)$, $D=12>0$ with $f_{xx}>0$, a single local minimum, so 'two local minima' is false. Also $\tfrac25+\tfrac23=\tfrac{16}{15}$, so (D) holds.

Step 5: Conclude.
The untrue statement is (C).
\[ \boxed{(C)} \]