Question

Consider the function \[ f(x,y)=(x-2)^2(y+3). \] Then

• A. \((2,-3)\) is not a stationary point of \(f\)
• B. \(f\) has a local minimum at \((2,-3)\)
• C. \(f\) has a local maximum at \((2,-3)\)
• D. \(f\) has neither a local maximum nor a local minimum at \((2,-3)\)

Hint:

If the second derivative test gives \[ \boxed{D=0,} \] the test is inconclusive. Always inspect the sign of the function around the stationary point. If it changes sign, the point is neither a maximum nor a minimum (typically a saddle-type point).

Answer 1

The Correct Option is D

Concept: To determine the nature of a point:
• Find the first partial derivatives and check whether the point is stationary.
• Apply the second derivative test. If it is inconclusive, inspect the behaviour of the function near the point.

Step 1: Find the first partial derivatives. Given, \[ f(x,y)=(x-2)^2(y+3). \] Differentiate with respect to \(x\): \[ f_x = 2(x-2)(y+3). \] Differentiate with respect to \(y\): \[ f_y = (x-2)^2. \] At the point \((2,-3)\), \[ f_x(2,-3)=0, \] and \[ f_y(2,-3)=0. \] Hence, \[ \boxed{(2,-3)\text{ is a stationary point}.} \]

Step 2: Find the second partial derivatives. \[ f_{xx}=2(y+3), \] \[ f_{yy}=0, \] \[ f_{xy}=2(x-2). \] At \((2,-3)\), \[ f_{xx}=0,\qquad f_{yy}=0,\qquad f_{xy}=0. \] Thus, \[ D = f_{xx}f_{yy}-(f_{xy})^2 = 0. \] The second derivative test is inconclusive.

Step 3: Examine the function near the point. Since \[ f(x,y) = (x-2)^2(y+3), \] note that \[ (x-2)^2\ge0. \] If \[ y>-3, \] then \[ y+3>0 \quad\Rightarrow\quad f(x,y)>0. \] If \[ y<-3, \] then \[ y+3<0 \quad\Rightarrow\quad f(x,y)<0. \] Thus, in every neighbourhood of \((2,-3)\), the function takes both positive and negative values. Hence, \[ \boxed{ (2,-3) \text{ is neither a local maximum nor a local minimum.} } \] Therefore, the correct option is \[ \boxed{(D).} \]