Max min problems.

[Studyinghard]

Find the point on the parabola y = x^2 that is closest to the point (3,0)

[brightsky]

Let

The shortest distance is the distance of the perpendicular line going from the point (3,0) to a particular point on the parabola. (Probably not the best way to put it though xD)

Let the point (3,0) be A and the point B be a point on the parabola such that AB is perpendicular to the tangent to the parabola at point B.

The gradient of the tangent to the parabola at the point B can be found with:

...(1)

Because AB is perpendicular to this line:

The gradient of AB is ....(2)

From (2):









Hence the only solution is m = 1.

So the point on the parabola is .

[matrix]

another approach:
the equation of the tangent to the parabola is y =2x +c,
so gradient of the normal is -1/2.
find the equaton of the line with m=-1/2 and passing (3,0)
fin the point of intersection between the parabola y = x^2 and the normal y=-1/2 (x-3)
one solution is x=1, other is x=-3/2 which is further from (3,0),
sub x=1 to find y
(1,1) answer
hope this helps.

[moekamo]

distance between two points:

so

max when

which is the same equation brightsky solved for x=1 so y=1 too

point is (1,1)

so many ways to do these q's :S

[naved_s9994]

similar to how its above