Let fsx, yd − x 2 1 y 2 2 2x 2 6y 1 14. Then fxsx, yd − 2x 2 2 fysx, yd − 2y 2 6 These partial derivatives are equal to 0 when x − 1 and y − 3, so the only critical point is s1, 3d. By completing the square, we find that fsx, yd − 4 1 sx 2 1d 2 1 sy 2 3d 2 Since sx 2 1d 2 > 0 and sy 2 3d 2 > 0, we have fsx, yd > 4 for all values of x and y. Therefore fs1, 3d − 4 is a local minimum, and in fact it is the absolute minimum of f. This can be confirmed geometrically from the graph of f, which is the paraboloid with vertex s1, 3, 4d shown in Figure 2.