Problem 11 - College Algebra and Calculus: An Applied Approach

Question

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

Step-by-Step Solution

Answer

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

Step 1: 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).

Step 2: 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 \( f_{xy} = 0 \).

Step 3: 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.