Problem 6 - APEX Calculus

Question

Find the critical points of the given function. Use the Second Derivative Test to determine if each critical point corresponds to a relative maximum, minimum, or saddle point. $$ f(x, y)=x^{2}+4 x+y^{2}-9 y+3 x y $$

Step-by-Step Solution

Answer

The only critical point is \((7, -6)\), which is a saddle point.

Step 1: Calculating Partial Derivatives

To find the critical points, we first calculate the partial derivatives of the function with respect to both variables. Start by finding \( f_x \) and \( f_y \).\[ f_x = \frac{\partial}{\partial x}(x^2 + 4x + y^2 - 9y + 3xy) = 2x + 4 + 3y \]\[ f_y = \frac{\partial}{\partial y}(x^2 + 4x + y^2 - 9y + 3xy) = 2y - 9 + 3x \]

Step 2: Finding Critical Points

Critical points occur where both partial derivatives are equal to zero. Solve the system of equations:1. \( 2x + 4 + 3y = 0 \)2. \( 2y - 9 + 3x = 0 \)Solving these:From equation 1:\[ 2x + 3y = -4 \]From equation 2:\[ 3x + 2y = 9 \]Multiply the first equation by 2 and the second by 3 to eliminate \(y\):\[ 4x + 6y = -8 \]\[ 9x + 6y = 27 \]Subtract the first from the second:\[ 5x = 35 \]\[ x = 7 \]Substitute \( x = 7 \) back into the first equation:\[ 2(7) + 3y = -4 \]\[ 14 + 3y = -4 \]\[ 3y = -18 \]\[ y = -6 \]The critical point is \((7, -6)\).

Step 3: Calculating Second Partial Derivatives

For the Second Derivative Test, compute the second partial derivatives:\[ f_{xx} = \frac{\partial^2}{\partial x^2}(f) = 2 \]\[ f_{yy} = \frac{\partial^2}{\partial y^2}(f) = 2 \]\[ f_{xy} = \frac{\partial^2}{\partial x \partial y}(f) = 3 \]

Step 4: Applying the Second Derivative Test

Use the second derivative test by calculating the determinant \( D \), where:\[ D = f_{xx}f_{yy} - (f_{xy})^2 \]Substitute the calculated values:\[ D = (2)(2) - (3)^2 = 4 - 9 = -5 \]Since \( D < 0 \), the point \((7, -6)\) is a saddle point.

Key Concepts

Partial Derivatives

When analyzing functions with multiple variables, partial derivatives are essential in finding critical points. Simply put, a partial derivative measures how the function changes as only one of the variables changes while holding the others constant.
For the function \(f(x, y) = x^2 + 4x + y^2 - 9y + 3xy\), the first step is to take the partial derivatives with respect to both \(x\) and \(y\).

• The partial derivative with respect to \(x\) is denoted as \(f_x\), and tells us how the function changes as \(x\) changes, resulting in \(f_x = 2x + 4 + 3y\).
• The partial derivative with respect to \(y\), denoted as \(f_y\), indicates how \(f\) changes as \(y\) changes, which gives us \(f_y = 2y - 9 + 3x\).

Once these derivatives are calculated, they're set to zero to find critical points, as these points occur where the slope in each direction is zero.

Second Derivative Test

The second derivative test is a powerful tool for classifying critical points found using partial derivatives. The test uses the second partial derivatives to determine whether a critical point is a relative maximum, minimum, or saddle point.To apply this test, calculate:

• \( f_{xx} \) and \( f_{yy} \), the second partial derivatives of \(f\) with respect to \(x\) and \(y\), respectively.
• \( f_{xy} \), which is the mixed partial derivative, showing how \(f\) changes with respect to both variables simultaneously.

For the function given:

• \( f_{xx} = 2 \)
• \( f_{yy} = 2 \)
• \( f_{xy} = 3 \)

With these derivatives, calculate the determinant \( D = f_{xx} f_{yy} - (f_{xy})^2 \). This expression helps assess the nature of the critical points.

Saddle Point

A saddle point is a type of critical point that isn't a local maximum or minimum. Instead, it's a point where the surface curves in different directions, like a saddle.
When the second derivative test is applied:

• If the determinant \(D < 0\), as in our problem where \( D = -5 \), the critical point is identified as a saddle point.
• This means that at the point, the surface does not have a crest or a trough, but rather shifts direction.

In the exercise, the critical point \((7, -6)\) was found, and since \(D\) is negative, this point is a saddle point. Such points are fascinating because they defy simple categorization as just peaks or valleys, offering a more complex landscape.

Relative Maximum and Minimum

Finding relative maximums and minimums involves analyzing critical points where \(D > 0\). These are points where the function reaches a local high or low within a region.

• If \( f_{xx} > 0 \) at the critical point and \(D > 0\), the point is a relative minimum.
• If \( f_{xx} < 0 \) and \(D > 0\), it's a relative maximum.

In this exercise, because the determinant \(D\) was negative, there were no relative minima or maxima. However, understanding these concepts is crucial because it helps differentiate whether the critical points are potential peaks or troughs within the surface plotted by the function. Such information is highly useful in optimization problems and in studying the behavior of multivariable functions.