Problem 1 - The Calculus with Analytic Geometry

Question

In Exercises 1 through 6, determine the relative extrema of \(f\), if there are any. \(f(x, y)=18 x^{2}-32 y^{2}-36 x-128 y-110\)

Step-by-Step Solution

Answer

The function has a saddle point at (1, -2).

Step 1: Find the First Partial Derivatives

Compute the first partial derivatives of the function with respect to both variables. For \ x: \(f_x(x, y) = \frac{\partial}{\partial x}(18x^2 - 32y^2 - 36x - 128y - 110) = 36x - 36\). For \ y: \(f_y(x, y) = \frac{\partial}{\partial y}(18x^2 - 32y^2 - 36x - 128y - 110) = -64y - 128\).

Step 2: Set Partial Derivatives to Zero

Set each first partial derivative to zero and solve for x and y to find the critical points. For \ x: \(36x - 36 = 0\) leads to \(x = 1\). For \ y: \(-64y - 128 = 0\) leads to \(y = -2\).

Step 3: Compute the Second Partial Derivatives

Find the second partial derivatives to use for the second derivative test. \(f_{xx}(x,y) = \frac{\partial}{\partial x} (36x - 36) = 36\). \(f_{yy}(x,y) = \frac{\partial}{\partial y} (-64y - 128) = -64\). \(f_{xy}(x,y) = \frac{\partial}{\partial y} (36x - 36) = 0\).

Step 4: Compute the Determinant of the Hessian Matrix

The Hessian matrix is given by: \(H = \begin{pmatrix} f_{xx} & f_{xy} \ f_{xy} & f_{yy} \end{pmatrix} \). Calculate its determinant: \(D = f_{xx} f_{yy} - (f_{xy})^2 = 36(-64) - (0)^2 = -2304\).

Step 5: Determine the Nature of the Critical Point

Since the determinant of the Hessian matrix \(D\) is negative (\(-2304\)), the critical point \((1, -2)\) is a saddle point.

Key Concepts

partial derivatives

Partial derivatives are used to understand how a function changes as its variables change. In this exercise, we have a function of two variables,\( f(x, y) = 18x^2 - 32y^2 - 36x - 128y - 110 \). To find the relative extrema, we first compute the partial derivatives with respect to both \( x \) and \( y \).
For \( x \):\( f_x(x, y) = \frac{\partial}{\partial x}(18x^2 - 32y^2 - 36x - 128y - 110) = 36x - 36 \).
For \( y \):\( f_y(x, y) = \frac{\partial}{\partial y}(18x^2 - 32y^2 - 36x - 128y - 110) = -64y - 128 \).
These derivatives tell us how the function changes with respect to small changes in \( x \) and \( y \), which is essential for finding critical points.

Hessian matrix

The Hessian matrix is a square matrix of second-order partial derivatives of a function. It helps in determining the curvature of the function. For our function, we computed the second partial derivatives:
\( f_{xx}(x, y) = \frac{\partial^2}{\partial x^2}(f) = 36 \)
\( f_{yy}(x, y) = \frac{\partial^2}{\partial y^2}(f) = -64 \)
\( f_{xy}(x, y) = \frac{\partial^2}{\partial x \partial y}(f) = 0 \).
Using these, the Hessian matrix is:
\[ H = \begin{pmatrix} f_{xx} & f_{xy} \ f_{xy} & f_{yy} \end{pmatrix} = \begin{pmatrix} 36 & 0 \ 0 & -64 \end{pmatrix} \].
This matrix is vital for determining the type of critical points, whether they are minima, maxima, or saddle points.

saddle point

A saddle point is a critical point that is neither a local maximum nor a local minimum. Instead, it has properties of both. After finding our critical point \((1, -2)\), we use the second partial derivatives to assemble the Hessian matrix and evaluate its determinant:
\( D = f_{xx} f_{yy} - (f_{xy})^2 = 36(-64) - (0)^2 = -2304 \).
Since \( D \) is negative, the critical point \((1, -2)\) is classified as a saddle point. At this point, the function increases in some directions and decreases in others, which is characteristic of a saddle point.

second derivative test

The second derivative test helps determine the nature of critical points. It relies on the determinant of the Hessian matrix (\( D \)) and the signs of the second partial derivatives:\( f_{xx} \) and \( f_{yy} \). Here's how it works:

• If \( D > 0 \) and \( f_{xx} > 0 \), the critical point is a local minimum.
• If \( D > 0 \) and \( f_{xx} < 0 \), the critical point is a local maximum.
• If \( D < 0 \), the critical point is a saddle point.
• If \( D = 0 \), the test is inconclusive.

Applying this to our function, we found that \( D = -2304 \) which is less than zero. Therefore, the critical point \((1, -2)\) is a saddle point, not a maximum or minimum.