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 first order partial derivatives.
Given
\[ f(x,y)=xy(5x+3y-6) \] Expanding,
\[ f(x,y)=5x^2y+3xy^2-6xy \] Now,
\[ f_x=10xy+3y^2-6y \] \[ f_x=y(10x+3y-6) \] and
\[ f_y=5x^2+6xy-6x \] \[ f_y=x(5x+6y-6) \]

Step 2: Find the critical points.
For critical points,
\[ f_x=0 \] and
\[ f_y=0 \] So,
\[ y(10x+3y-6)=0 \] and
\[ x(5x+6y-6)=0 \] This gives the critical points
\[ (0,0),\left(\frac{6}{5},0\right),(0,2),\left(\frac{2}{5},\frac{2}{3}\right) \] Hence, \(f\) has exactly \(4\) critical points.
Therefore, option (A) is true.

Step 3: Use second derivative test.
Now,
\[ f_{xx}=10y \] \[ f_{yy}=6x \] \[ f_{xy}=10x+6y-6 \] The Hessian determinant is
\[ D=f_{xx}f_{yy}-(f_{xy})^2 \]

Step 4: Classify the critical points.
At \((0,0)\),
\[ D=0-( -6)^2=-36<0 \] So, \((0,0)\) is a saddle point.
At \(\left(\frac{6}{5},0\right)\),
\[ D=0-(6)^2=-36<0 \] So, \(\left(\frac{6}{5},0\right)\) is a saddle point.
At \((0,2)\),
\[ D=0-(6)^2=-36<0 \] So, \((0,2)\) is a saddle point.
At \(\left(\frac{2}{5},\frac{2}{3}\right)\),
\[ f_{xx}=\frac{20}{3} \] and
\[ D=12>0 \] Since \(D>0\) and \(f_{xx}>0\), this point is a local minimum.
Thus, there is only one local minimum, not two.

Step 5: Check option (D).
For the critical point
= \[ (a,b)=\left(\frac{2}{5},\frac{2}{3}\right) \] we get
\[ a+b=\frac{2}{5}+\frac{2}{3} \] \[ a+b=\frac{6+10}{15} \] \[ a+b=\frac{16}{15} \] Therefore, option (D) is true.

Step 6: Final conclusion.
The statement which is NOT true is
\[ \boxed{(C)} \]