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Chapter 13: Problem 5

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

Short Answer

Expert verified
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 by step solution

01

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 \( f_y = 2(y - 3) \).
02

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.
03

Find Second Derivatives

To find the second derivatives, differentiate \( f_x \) and \( f_y \) with respect to x and y respectively. This gives: \(f_{xx} = 2\), \(f_{xy} = 0\), \(f_{yy} = 2\).
04

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.

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