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# Find the point of the graph nearest to 0

Find the points on the surface

``` z = 2xy + 1 / 4
```

nearest to the origin.

Review exercise. Find the points on the curve

``` f(x) = x2 + 2x + 2
```

to the 0. Now find the partial derivatives:

``` ∂f / dx = 2y = 0,  ∂d / dy = 2x = 0,
```

solving

``` x = 0, y = 0 and z = 1 / 4,
```

and hence

``` the point nearest to the origin is ( 0, 0, 1 / 4 ).
```

Continue. Find the points on the surface

``` z = 2xy + 1 / 4
```

nearest to the origin. Note that this is different from finding the minimum of

``` z = f( x, y ) = 2xy + 1 / 4.
```

Q: Point ( x, y, z ), distance to the 0-point is

``` g( x, y, z ) = x2 + y2 + z2,
```

where z = 2xy + 1 / 4. Then minimize

``` h( x, y ) = x2 + y2 + ( 2xy + 1 / 4 )2,
```

i.e. the square of the distance of a point ( x, y, z ) on the surface to 0.

Solution:

``` h( x, y ) = x2 + y2 + 4 x2 y2 + xy + 1 / 16.
```

Then

``` ∂h / ∂x = 2x + 8xy2 + y = 0,
∂h / ∂y = 2y + 8x2y + x = 0,
```

hence

``` x = 0, y = 0. (Why?)
```

Find all x, y satisfying the system

``` 2x2 + 8x2y2 + xy = 0
2y2 + 8x2y2 + xy = 0
----------------------
2x2 - 2y2 = 0
```

hence

``` x = ± y.
```

Substitute back ( x = y ):

``` 2x + 8x3 + x = 0
8x3 + 3x = 0
x ( 8x2 + 3 ) = 0,
x = 0 → y = 0.

(1) → x = -y / ( 2 + 8 y2 )
(2) 2 y + y2 / ( 2 + 8y2 )2 + ( -y ) / ( 2+ 8y2 ) = 0
```

Exercise. Let

``` y = x2 - a.
```

Find the nearest point to 0.

``` d(x) = x2 + y2
= x2 + x4 - 2ax2 + a2,
```

``` d′(x) = 2x + 4x3 - 4ax = x ( 2 - 4a + 4x2 ) = 0
→ x = 0 or 2 - 4a + 4x2 = 0,
```

hence

``` x = 0 or x = ± ( a - 1 / 2 )1/2 ∈ ℝ if a > 1 / 2:
Case 1: -( a - 1 / 2 )1/2 -( a - 1 / 2 )1/2 if a > 1 / 2,
Case 2: if a ≤ 1 / 2
If a = 1 / 2, d′(x) = x3.
```

As a passes through 1/2, one solution (which is also the minimum of f) becomes two solutions (bifurcation).