Problem 335 - Calculus 3

Question

For the following exercises, use the second derivative test to identify any critical points and determine whether each critical point is a maximum, minimum, saddle point, or none of these. $$ f(x, y)=x^{2}+10 x y+y^{2} $$

Step-by-Step Solution

Answer

The critical point (0, 0) is a saddle point.

Step 1: Find the first partial derivatives

First, we find the partial derivatives of the function. The partial derivative of \( f \) with respect to \( x \) is:\[ f_x(x, y) = 2x + 10y \]The partial derivative of \( f \) with respect to \( y \) is:\[ f_y(x, y) = 10x + 2y \]

Step 2: Set partial derivatives to zero to find critical points

To find the critical points, set the first partial derivatives equal to zero:\[ 2x + 10y = 0\]\[ 10x + 2y = 0\]

Step 3: Solve the system of equations

Solve the system of equations to find the critical point.From the first equation: \( 2x + 10y = 0 \), we get \( x = -5y \).Substitute \( x = -5y \) into the second equation: \[ 10(-5y) + 2y = 0 \] \[ -50y + 2y = 0 \] \[ -48y = 0 \] So, \( y = 0 \). Substituting \( y = 0 \) into \( x = -5y \) gives \( x = 0 \).Thus, the critical point is \((0, 0)\).

Step 4: Compute the second partial derivatives

Next, compute the second partial derivatives to use in the second derivative test. These are:\[ f_{xx} = 2, \quad f_{yy} = 2, \quad f_{xy} = 10 \]

Step 5: Calculate the determinant of the Hessian matrix

The Hessian matrix \( H \) is given by:\[ H = \begin{bmatrix} f_{xx} & f_{xy} \ f_{xy} & f_{yy} \end{bmatrix} = \begin{bmatrix} 2 & 10 \ 10 & 2 \end{bmatrix} \]The determinant of \( H \), denoted by \( D \), is calculated as:\[ D = f_{xx}f_{yy} - (f_{xy})^2 = (2)(2) - (10)^2 = 4 - 100 = -96 \]

Step 6: Apply the second derivative test

Since the determinant \( D = -96 \) is less than zero, the critical point \((0, 0)\) is a saddle point, according to the second derivative test. No maximum or minimum at this critical point.

Key Concepts

Critical Points

In the realm of calculus, when analyzing functions of several variables, identifying critical points is a crucial step towards understanding the behavior of the function. Critical points occur where the first partial derivatives of a function are zero. For the function given, we found the critical point by solving the system of equations derived from setting the partial derivatives to zero:

• The equation from the derivative with respect to \(x\): \(2x + 10y = 0\)
• The equation from the derivative with respect to \(y\): \(10x + 2y = 0\)

By substituting and solving these equations, we discovered the critical point at \((0, 0)\). This point potentially represents extremes like a local maximum or minimum, or a saddle point.
It's important to note that finding a critical point does not immediately categorize it as a max, min, or saddle point. Further testing is needed, which brings us to our next tool: the Hessian Matrix and Second Derivative Test.

Partial Derivatives

Partial derivatives extend the concept of derivatives to functions of multiple variables. They measure the rate at which the function changes as one of the variables is varied, keeping the others constant.
For our function \(f(x, y) = x^2 + 10xy + y^2\), we calculated the partial derivatives, which serve to find directions of change:

• \(f_x(x, y) = 2x + 10y\): This shows how \(f\) changes with \(x\).
• \(f_y(x, y) = 10x + 2y\): This shows how \(f\) changes with \(y\).

Setting these partial derivatives to zero, we identify potential sites of local extremes, known as critical points. Here, calculating these derivatives helps us navigate the landscape of the function, like a topographical survey of a landscape, to pinpoint influential locations for further analysis.

Hessian Matrix

The Hessian Matrix is a square matrix composed of the second partial derivatives of a function. It provides invaluable information regarding the curvature of the function around critical points, similar to analyzing how steep or flat a hill is in every direction around a given point.
For our function, the Hessian Matrix \(H\) was constructed as:

• \(f_{xx} = 2\)
• \(f_{yy} = 2\)
• \(f_{xy} = f_{yx} = 10\)
• Leading to \(H = \begin{bmatrix} 2 & 10 \ 10 & 2 \end{bmatrix}\)

This matrix is pivotal because the determinant of the Hessian, denoted as \(D\), helps classify the nature of the critical point:
\[D = f_{xx}f_{yy} - (f_{xy})^2 = (2)(2) - 10^2 = 4 - 100 = -96\]When \(D\) is less than zero, as in this case, the critical point is a saddle point.

Saddle Point

A saddle point is a critical point that indicates a change in curvature, somewhat resembling a saddle in topographical terms. It is neither a local maximum nor a minimum but a point where different directions on the surface curve upwards and downwards.
In our exercise, the second derivative test revealed a negative Hessian determinant \(D = -96\). Such a result confirms that the critical point \((0, 0)\) is a saddle point.

• Saddle points reflect instability, as they are points where the surface "saddles" rather than displaying consistent curvature in any direction.
• Because the determinant is negative, it suggests that on one axis, the surface curves upwards, and on another, it curves downwards.

Understanding saddle points is essential in optimization and in fields such as economics and physics, where such unstable equilibrium points bear interesting implications.