Question

Question #3d707

Answer 1

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.