7. Find the critical points of the given function. Use the Second Derivative Test to determine if each critical point corresponds to a relative maximum, minimum, or saddle point.

\(f(x, y) = x^2 + 3y^2 − 6y + 4xy\)

  1. Find the partial derivatives

\(f_x(x, y) = 2x + 4y\)

\(f_y(x, y) = 6y − 6 + 4x\)

  1. Seƫ each derivitive equal to 0 and solve for x and y

\(2x + 4y=0\)

\(6y − 6 + 4x=0\)

\(y=-\frac{1}{2}x\) \(6(-\frac{1}{2}x) − 6 + 4x=0\)

\(-3x-6+4x=0\)

\(x=6\)

\(y=-\frac{1}{2}*6=-3\)

There is one critical point: (6;-3)

  1. Find partial second derivitive

\(f_{xx}(x, y) = 2\)

\(f_{yy}(x, y) = 6\)

\(f_{xy}(x, y) = 4\)

\(f_{xy}(x, y) = 4\)

  1. Calculate D

\(D = f_{xx}(x_0, y_0)+f_{yy}(x_0, y_0) − f_{xy}^2(x_0, y_0)\)

\(D = 2+6 − 4^2=8-16=-8\)

If \(D<0\) the critical point of (6;-3) is a saddle point