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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) = x^{2}+ 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 ) = x^{2}+ y^{2}+ z^{2},

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

h( x, y ) = x^{2}+ y^{2}+ ( 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 ) = x^{2}+ y^{2}+ 4 x^{2}y^{2}+ xy + 1 / 16.

Then

∂h / ∂x = 2x + 8xy^{2}+ y = 0, ∂h / ∂y = 2y + 8x^{2}y + x = 0,

hence

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

Find all x, y satisfying the system

2x^{2}+ 8x^{2}y^{2}+ xy = 0 2y^{2}+ 8x^{2}y^{2}+ xy = 0 ---------------------- 2x^{2}- 2y^{2}= 0

hence

x = ± y.

Substitute back ( x = y ):

2x + 8x^{3}+ x = 0 8x^{3}+ 3x = 0 x ( 8x^{2}+ 3 ) = 0, x = 0 → y = 0. (1) → x = -y / ( 2 + 8 y^{2}) (2) 2 y + y^{2}/ ( 2 + 8y^{2})^{2}+ ( -y ) / ( 2+ 8y^{2}) = 0

**Exercise.** Let

y = x^{2}- a.

Find the nearest point to 0.

d(x) = x^{2}+ y^{2}= x^{2}+ x^{4}- 2ax^{2}+ a^{2},

d′(x) = 2x + 4x^{3}- 4ax = x ( 2 - 4a + 4x^{2}) = 0 → x = 0 or 2 - 4a + 4x^{2}= 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) = x^{3}.

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