Question #3d707

1 Answer
Jan 16, 2018

Answer:

This is an elliptic paraboloid, having a single minimum.

Explanation:

Let #f(x, y) = 2x^2 + 2y^2 + 1#.
Then #f_x(x, y) = 4x#
and #f_y(x, y) = 4y#.
Clearly #f_x = 0# when x = 0, and #f_y = 0# when y = 0.
The two partial derivatives are zero only at (0, 0).

This is the only critical point.

Now, we could the second partials test.
#f_(xx)(x, y) = 4# (Second partial with respect to x twice.)
#f_(xy)(x, y) = 0 = f_(yx)(x, y)#
#f_(yy)(x, y) = 4#
Therefore # D = f_(xx)(x, y)f_(yy)(x, y) - f_(xy)(x, y)f_(yx)(x, y)#
#= 16#
Since D > 0, we examine f-xx. Since f-xx = 4 > 0, this is a (local) minimum.

However, we might also skip the partials test. We observe that since both variables are squared, the global minimum value must occur where x = 0 and y = 0. That is, the minimum value is #0 + 0 + 1 = 1#, and the origin is the minimum.