Problem 5 - College Algebra and Calculus: An Applied Approach

Question

Examine the function for relative extrema and saddle points. $$ f(x, y)=(x-1)^{2}+(y-3)^{2} $$

Step-by-Step Solution

Answer

The function \(f(x, y)=(x-1)^{2}+(y-3)^{2}\) has a relative minimum at the point (1,3). It has no saddle points.

Step 1: Find the First Derivatives

The first derivatives of the function are determined using the basic rules of differentiation. The first derivatives are \( f_x = 2(x - 1) \) and \( f_y = 2(y - 3) \).

Step 2: Set First Derivatives to Zero and Solve

Set the first derivatives equal to zero to find potential extremas or saddle points: \(f_x = 2(x - 1) = 0 \) and \(f_y = 2(y - 3) = 0\). Solving these gives \(x = 1\) and \(y = 3\). So, (1,3) is a critical point.

Step 3: Find Second Derivatives

To find the second derivatives, differentiate \( f_x \) and \( f_y \) with respect to x and y respectively. This gives: \(f_{xx} = 2\), \(f_{xy} = 0\), \(f_{yy} = 2\).

Step 4: Apply Second Derivative Test

The second derivative test involves evaluating \(D = f_{xx}*f_{yy} - (f_{xy})^2\) at the point (1, 3). Calculating this gives \(D = 2*2 - 0 = 4\), which is greater than 0. And since \(f_{xx}\) is also greater than 0, the point (1,3) is a relative minimum.