on the surface
z=2⋅y^2+2⋅x^2
the closest point to the plain:
is: (4/1 , 1/4 , 5/4)



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

Explore how to find the closest point on the surface z = 2y^2 + 2x^2 to the point (4, 1/4, 5/4) using distance minimization techniques. This problem involves advanced concepts in distance calculation and surface equations.

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