Problem 9 - Applied Calculus

Question

Find all the critical points and determine whether each is a local maximum, local minimum, a saddle point, or none of these. $$f(x, y)=y^{3}-3 x y+6 x$$

Step-by-Step Solution

Answer

The critical point (4, 2) is a saddle point.

Step 1: Find the first partial derivatives

To find the critical points, first compute the partial derivatives of the function \( f(x, y) = y^3 - 3xy + 6x \) with respect to \( x \) and \( y \).\(\frac{\partial f}{\partial x} = -3y + 6\)\(\frac{\partial f}{\partial y} = 3y^2 - 3x\)

Step 2: Set the first partial derivatives to zero

Set the partial derivatives found in Step 1 to zero to find the critical points.\(-3y + 6 = 0\) \\(3y^2 - 3x = 0\)Solving the first equation:\( y = 2 \)Plug \( y = 2 \) into the second equation:\(3(2)^2 - 3x = 0 \12 - 3x = 0 \x = 4\)

Step 3: Identify critical point(s)

From the previous step, we found one critical point:\((x, y) = (4, 2)\)

Step 4: Compute the second partial derivatives

Now, calculate the second partial derivatives to use the second derivative test:\(\frac{\partial^2 f}{\partial x^2} = 0 \\frac{\partial^2 f}{\partial y^2} = 6y \\frac{\partial^2 f}{\partial x \partial y} = -3\)

Step 5: Evaluate the second derivative test

Use the second derivative test given by the determinant of the Hessian matrix:The Hessian matrix is:\[H = \begin{bmatrix} \frac{\partial^2 f}{\partial x^2} & \frac{\partial^2 f}{\partial x \partial y} \ \frac{\partial^2 f}{\partial y \partial x} & \frac{\partial^2 f}{\partial y^2} \end{bmatrix} = \begin{bmatrix} 0 & -3 \ -3 & 6y \end{bmatrix}\]Calculate the determinant at the critical point \((4, 2)\):\(D = \begin{vmatrix} 0 & -3 \ -3 & 6(2) \end{vmatrix} = 0 \cdot 12 - (-3)(-3) = -9\)Since \(D < 0\), the critical point \((4, 2)\) is a saddle point.

Key Concepts

Partial Derivatives

Partial derivatives help us analyze functions of multiple variables by looking at how the function changes when only one variable is varied while others are held constant. In the function provided, \(f(x, y) = y^3 - 3xy + 6x\), we need to find two partial derivatives: one with respect to \(x\) and one with respect to \(y\).

• Partial derivative with respect to \(x\): Here, differentiate only looking at \(x\) while treating \(y\) as a constant. Calculating gives \( \frac{\partial f}{\partial x} = -3y + 6 \).
• Partial derivative with respect to \(y\): Here, differentiate considering \(y\) while \(x\) remains constant. This results in \( \frac{\partial f}{\partial y} = 3y^2 - 3x \).

These derivatives help determine the rate of change in each direction and are vital for locating critical points.

Second Derivative Test

The second derivative test is a method used in calculus to classify critical points as potential local maxima, minima, or saddle points. Once the first derivatives are zero, indicating a critical point, second derivatives provide further classification clues.
For function \(f\), you'll move to calculate the second partial derivatives:

• \(\frac{\partial^2 f}{\partial x^2} = 0\)
• \(\frac{\partial^2 f}{\partial y^2} = 6y\)
• \(\frac{\partial^2 f}{\partial x \partial y} = -3\)

These values form part of the Hessian matrix, a critical element of the test that helps you establish the nature of the critical points.

Hessian Matrix

The Hessian matrix is a square matrix of second-order partial derivatives of a function. It plays an important role in the second derivative test to help assess the nature of critical points. For our function, the Hessian matrix becomes:\[H = \begin{bmatrix} 0 & -3 \ -3 & 6y \end{bmatrix}\]At a critical point \((4, 2)\), calculate the determinant \(D\) of this matrix to examine the characteristics:\[D = \begin{vmatrix} 0 & -3 \ -3 & 12 \end{vmatrix} = 0 \times 12 - (-3)(-3) = -9\]If the determinant \(D\) is less than 0, as in this case (-9), the critical point is classified as a saddle point, indicating the function behaves like a saddle around this point.

Saddle Point Identifications

Saddle points are critical points where the function behaves differently in perpendicular directions, unlike maxima or minima. When using the Hessian matrix, if the determinant \(D\) is negative, it indicates a saddle point. In the case of the function \(f\), this involved calculating the Hessian determinant at \((4, 2)\) and finding it to be \(-9\). This negative value clearly identifies our point as a saddle point. Such points are neither local maxima nor minima and represent a point where paths flowing out of the point won't converge or diverge consistently.