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Chapter 13: Problem 11

Examine the function for relative extrema and saddle points. $$ f(x, y)=3 x^{2}+2 y^{2}-12 x-4 y+7 $$

Short Answer

Expert verified
The function \( f(x, y)=3x^{2}+2y^{2}-12x-4y+7 \) has a relative minimum at the point (2,1).

Step by step solution

01

Find the first order partial derivatives and the critical points

Firstly, we need to find the first order partial derivatives of \( f \). The partial derivative with respect to \( x \), \( f_{x} \), and the partial derivative with respect to \( y \), \( f_{y} \), are as follows: \n \( f_{x} = 6x - 12 \) \n \( f_{y} = 4y - 4 \). \n Finding the critical points involves setting \( f_{x} = 0 \) and \( f_{y} = 0 \). Solving these equations for \( x \) and \( y \) respectively gives us the critical points (2,1).
02

Calculate the second order partial derivatives

Next, calculate the second order partial derivatives: \( f_{xx} \), \( f_{yy} \) and \( f_{xy} \). These are given by: \n \( f_{xx} = 6 \) \n \( f_{yy} = 4 \) \n \( f_{xy} = 0 \).
03

Apply the Second Partial Derivative Test

This test involves evaluating \( D = f_{xx}f_{yy} - f_{xy}^{2} \) at the critical point (2,1). If \( D > 0 \) and both \( f_{xx} \) and \( f_{yy} \) are positive, we have a relative minimum. If \( D > 0 \) and both \( f_{xx} \) and \( f_{yy} \) are negative, we have a relative maximum. If \( D < 0 \), we have a saddle point. \n For our critical point (2,1), we find that \( D = 6 * 4 - 0^{2} = 24 > 0 \), and both \( f_{xx} \) and \( f_{yy} \) are positive. Therefore, (2,1) is a relative minimum.

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