Question

Consider the function \( f(x, y) = 5 - 4 \sin x + y^2 \) for \( 0<x<2\pi \) and \( y \in \mathbb{R} \). The set of critical points of \( f(x, y) \) consists of

• A. a point of local maximum and a point of local minimum
• B. a point of local maximum and a saddle point
• C. a point of local minimum and a saddle point
• D. a point of local maximum, a point of local minimum and a saddle point

Hint:

For critical points, set the first derivatives equal to zero, and use the second derivative test to classify them.

Answer 1

The Correct Option is C

Step 1: Finding the critical points. 
The critical points of a function occur when the first derivatives with respect to \( x \) and \( y \) are both zero. To find the critical points, we take the partial derivatives of \( f(x, y) \). 
\[ \frac{\partial f}{\partial x} = -4 \cos x \quad \text{and} \quad \frac{\partial f}{\partial y} = 2y. \] Setting these equal to zero gives: \[ -4 \cos x = 0 \quad \Rightarrow \quad \cos x = 0 \quad \Rightarrow \quad x = \frac{\pi}{2}, \frac{3\pi}{2}. \] \[ 2y = 0 \quad \Rightarrow \quad y = 0. \] Thus, the critical points occur at \( \left( \frac{\pi}{2}, 0 \right) \) and \( \left( \frac{3\pi}{2}, 0 \right) \). 
Step 2: Classification of critical points. 
To classify the critical points, we examine the second derivatives: \[ \frac{\partial^2 f}{\partial x^2} = 4 \sin x, \quad \frac{\partial^2 f}{\partial y^2} = 2, \quad \frac{\partial^2 f}{\partial x \partial y} = 0. \] At \( x = \frac{\pi}{2} \), \( \frac{\partial^2 f}{\partial x^2} = 4 \), which indicates a local minimum. At \( x = \frac{3\pi}{2} \), \( \frac{\partial^2 f}{\partial x^2} = -4 \), which indicates a saddle point. Additionally, the point \( \left( \frac{3\pi}{2}, 0 \right) \) is a local maximum. 

Step 3: Conclusion. 
Thus, the set of critical points consists of a point of local minimum and a saddle point.