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Chapter 13: Problem 10

Examine the function for relative extrema and saddle points. $$ f(x, y)=x^{2}+6 x y+10 y^{2}-4 y+4 $$

Short Answer

Expert verified
The function has a relative minimum at the point (-5/4, 5/6).

Step by step solution

01

Compute the first-order partial derivatives f_x and f_y

Take the partial derivatives of the given function \(f(x, y)\) with respect to \(x\) and \(y\). \[f_x(x, y) = 2x+6y\] \[f_y(x, y) = 6x+20y-4\] These partial derivatives represent the slope of the function in the \(x\) and \(y\) directions, respectively.
02

Find the critical points

Set these two equations to zero and solve them simultaneously, \[2x+6y=0\] \[6x+20y-4=0\] By solving these equations, we find the only critical point as \((-5/4, 5/6)\).
03

Compute the second-order partial derivatives

Now compute the second order partial derivatives, \[f_{xx}(x, y) = 2\] \[f_{yy}(x, y) = 20\] \[f_{xy}(x, y) = f_{yx}(x, y) = 6\] The mixed second order partial derivatives are the same, due to Clairaut's theorem.
04

Apply the second derivative test

The second derivative test involves evaluating the determinant of the Hessian matrix, \(D = f_{xx} * f_{yy} - f_{xy} ^2\), at the critical point. Substituting \(f_{xx} = 2\), \(f_{yy} = 20\) and \(f_{xy} = 6\), we get \(D = 2*20 - 6^2 = 40 - 36 = 4\). Since \(D > 0\) and \(f_{xx} > 0\), the critical point is a relative minimum.

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