Question

Let \( f : \mathbb{R}^2 \to \mathbb{R} \) be defined by \[ f(x, y) = e^{y}(x^2 + y^2) \quad \text{for all } (x, y) \in \mathbb{R}^2. \] Then, which one of the following is TRUE?

• A. The number of points at which \( f \) has a local minimum is 2
• B. The number of points at which \( f \) has a local maximum is 2
• C. The number of points at which \( f \) has a local minimum is 1
• D. The number of points at which \( f \) has a local maximum is 1

Hint:

To find local minima or maxima, first find the critical points by solving the first partial derivatives. Then, use the second derivative test to classify the critical points.

Answer 1

The Correct Option is C

Step 1: First and second derivatives of \( f(x, y) \).
The function is given by: \[ f(x, y) = e^{y}(x^2 + y^2). \] To find the critical points, compute the first partial derivatives: \[ \frac{\partial f}{\partial x} = 2x e^y, \quad \frac{\partial f}{\partial y} = e^y (2y + x^2). \] Now, set these equal to 0 to find critical points: \[ 2x e^y = 0 \quad \text{and} \quad e^y (2y + x^2) = 0. \] Since \( e^y \neq 0 \) for any \( y \), we get: \[ x = 0 \quad \text{and} \quad 2y + x^2 = 0. \] Thus, at \( x = 0 \), we have: \[ 2y = 0 \quad \Rightarrow y = 0. \] So, the only critical point is \( (x, y) = (0, 0) \). Step 2: Second derivative test.
To classify the critical point, compute the second partial derivatives: \[ f_{xx} = 2e^y, \quad f_{yy} = e^y(2 + x^2), \quad f_{xy} = 2x e^y. \] At \( (0, 0) \), we have: \[ f_{xx}(0, 0) = 2, \quad f_{yy}(0, 0) = 2, \quad f_{xy}(0, 0) = 0. \] The discriminant is: \[ D = f_{xx} f_{yy} - (f_{xy})^2 = 2 \times 2 - 0^2 = 4. \] Since \( D>0 \) and \( f_{xx}>0 \), the critical point \( (0, 0) \) is a local minimum. Final Answer: \[ \boxed{\text{The number of points at which } f \text{ has a local minimum is } 1.} \]