Question

Examine the function f(x, y) = 1 + sin (x² + y²) for extreme.

Answer 1

f(x, y) = 1 + sin(x² + y²) 

f_x = 2xcos (x² + y²)

f_y = 2ycos(x² + y²)

Now f_x = 0 and f_y = 0 implies

i.e., 2xcos(x² + y²) = 0 and 2ycos(x² + y²) = 0

∴ x = 0, y = 0 and (0, 0) is the stationary point.

A = f_xx = – 4x² sin (x² + y²) + 2cos (x² + y²)

B = f_xy = – 4xy sin (x² + y²)

 = f_yy = – 4y²sin(x² + y²) + 2cos(x² + y²)

At (0, 0); A = 2, B = 0, C = 2

∴ AC – B² = 4 > 0

Since, AC – B² > 0, A = 2 > 0, (0, 0) is minimum point and the minimum value of  f (x, y) = f (0, 0) = 1.