Question

Consider the function \( f(x, y) = x^2 + xy^2 + y^4 \), then which of the following statement is correct:

• A. \( f(x, y) \) has neither a maxima nor a minima at the origin (0,0)
• B. \( f(x, y) \) has a minimum value at the origin (0, 0)
• C. origin (0, 0) is a saddle point of \( f(x, y) \)
• D. \( f(x, y) \) has a maximum value at the origin (0, 0)

Hint:

When the second derivative test fails (\(D=0\)), don't give up! Look for a way to analyze the function's sign near the critical point. Completing the square is a powerful algebraic technique that can often reveal if the function is always non-negative or non-positive near the point.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
To determine the nature of a critical point for a function of two variables, we typically use the second partial derivative test. If this test is inconclusive (Discriminant D=0), we must analyze the function's behavior in the neighborhood of the critical point directly.

Step 2: Find Critical Points:
First, find the partial derivatives and set them to zero. \[ f_x = \frac{\partial f}{\partial x} = 2x + y^2 = 0 \] \[ f_y = \frac{\partial f}{\partial y} = 2xy + 4y^3 = 2y(x + 2y^2) = 0 \] From the second equation, either \( y=0 \) or \( x = -2y^2 \). Case 1: If \( y=0 \), the first equation gives \( 2x + 0 = 0 \implies x=0 \). So \((0,0)\) is a critical point. Case 2: If \( x = -2y^2 \), substitute into the first equation: \( 2(-2y^2) + y^2 = 0 \implies -4y^2 + y^2 = 0 \implies -3y^2 = 0 \implies y=0 \). This again gives \(x=0\). The only critical point is the origin \((0,0)\).

Step 3: Second Derivative Test:
Now, find the second partial derivatives: \[ f_{xx} = 2 \] \[ f_{yy} = 2x + 12y^2 \] \[ f_{xy} = 2y \] Evaluate at the critical point \((0,0)\): \[ f_{xx}(0,0) = 2, \quad f_{yy}(0,0) = 0, \quad f_{xy}(0,0) = 0 \] Calculate the discriminant \( D = f_{xx}f_{yy} - (f_{xy})^2 \): \[ D(0,0) = (2)(0) - (0)^2 = 0 \] Since \(D=0\), the second derivative test is inconclusive.

Step 4: Analyze the Function Directly:
We must investigate the behavior of \( f(x,y) \) near \((0,0)\). The value at the origin is \(f(0,0)=0\). Let's try to rewrite the function by completing the square: \[ f(x,y) = x^2 + xy^2 + y^4 = \left(x^2 + xy^2 + \frac{y^4}{4}\right) - \frac{y^4}{4} + y^4 = \left(x + \frac{y^2}{2}\right)^2 + \frac{3y^4}{4} \] The first term, \( \left(x + \frac{y^2}{2}\right)^2 \), is a square, so it is always greater than or equal to 0. The second term, \( \frac{3y^4}{4} \), is also always greater than or equal to 0. The sum of two non-negative terms must be non-negative. Therefore, \( f(x,y) \ge 0 \) for all \( (x,y) \). Since \(f(0,0) = 0\), the function has a global (and local) minimum at the origin.