To find the closest point on the surface $z = 2y^2 + 2x^2$ to the point $\left( \frac{4}{1} , \frac{1}{4} , \frac{5}{4} \right)$, we need to minimize the distance between any point $(x, y, z)$ on the surface and this given point.

To clarify, the given point is $(4, \frac{1}{4}, \frac{5}{4})$. The surface equation is: $z = 2y^2 + 2x^2$

We are looking for the point $(x, y, z)$ on this surface that minimizes the distance to $(4, \frac{1}{4}, \frac{5}{4})$.

The distance $D$ between the points $(x, y, z)$ and $(4, \frac{1}{4}, \frac{5}{4})$ is given by: $D = \sqrt{(x - 4)^2 + \left(y - \frac{1}{4}\right)^2 + \left(z - \frac{5}{4}\right)^2}$

Since $z$ must satisfy the surface equation $z = 2y^2 + 2x^2$, we substitute $z$ into the distance formula: $D = \sqrt{(x - 4)^2 + \left(y - \frac{1}{4}\right)^2 + \left(2x^2 + 2y^2 - \frac{5}{4}\right)^2}$

We need to minimize this distance $D$. Instead of minimizing $D$, we can minimize $D^2$ to simplify our calculations. Therefore, we minimize: [ D^2 = (x - 4)^2 + \left(y - \frac{1}{4}\right)^2 + \left(2x^2 + 2y^2 - \frac{5