# Question #3d707

Jan 16, 2018

This is an elliptic paraboloid, having a single minimum.

#### Explanation:

Let $f \left(x , y\right) = 2 {x}^{2} + 2 {y}^{2} + 1$.
Then ${f}_{x} \left(x , y\right) = 4 x$
and ${f}_{y} \left(x , y\right) = 4 y$.
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.
${f}_{\times} \left(x , y\right) = 4$ (Second partial with respect to x twice.)
${f}_{x y} \left(x , y\right) = 0 = {f}_{y x} \left(x , y\right)$
${f}_{y y} \left(x , y\right) = 4$
Therefore $D = {f}_{\times} \left(x , y\right) {f}_{y y} \left(x , y\right) - {f}_{x y} \left(x , y\right) {f}_{y x} \left(x , y\right)$
$= 16$
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.