Problem 10 - College Algebra and Calculus: An Applied Approach

Question

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

Step-by-Step Solution

Answer

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

Step 1: 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.

Step 2: 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 \((-5/4, 5/6)\).

Step 3: 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.

Step 4: 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.