Problem 6 - College Algebra and Calculus: An Applied Approach

Question

Examine the function for relative extrema and saddle points. $$ f(x, y)=9-(x-3)^{2}-(y+2)^{2} $$

Step-by-Step Solution

Answer

The critical point of the function \(f(x, y)=9-(x-3)^{2}-(y+2)^{2}\) is found to be at (3,-2). After evaluating the determinant of the Hessian matrix, it is determined that this is a relative maximum.

Step 1: Find Partial Derivatives

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

Step 2: Set Partial Derivatives Equal to Zero

Then, set these equal to zero to solve for \(x\) and \(y\) respectively. This will give you coordinates of the critical points. \[-2(x-3) = 0 \rightarrow x = 3,\] \[ -2(y+2) = 0 \rightarrow y = -2.\] Thus, the critical point is at (3, -2).

Step 3: Evaluate the Second Partial Derivatives

After finding the critical point, find the second partial derivatives:\[\frac{\partial^2 f}{\partial x^2} = -2,\] \[ \frac{\partial^2 f}{\partial y^2} = -2,\] and the mixed second partial derivative: \[\frac{\partial^2 f}{\partial y \partial x} = 0.\]

Step 4: Evaluate the Determinant of the Hessian Matrix

Now, find the determinant of the Hessian matrix. It is defined as \[ H = \frac{\partial^2 f}{\partial x^2} \cdot \frac{\partial^2 f}{\partial y^2} - \left(\frac{\partial^2 f}{\partial y \partial x}\right)^2.\] Plug in the values to get that \(H = (-2)(-2) - (0)^2 = 4.\] If the determinant is positive, the point will be an extremum, so our point (3,-2) is a relative extremum.

Step 5: Checking whether it is a relative maximum or minimum

Examine the sign of the second-order partial derivatives. If both are negative, then our point is a local maximum. If both are positive, then it is a local minimum. Since both second partial derivatives are negative, our point is a relative maximum.