What is the shortest distance from the surface xy+12x+z^2=129 to the origin?

Answer 1

There are a couple ways of doing it, both involving gradients.

So I assume that you have covered gradients, but I will not assume that you learned about

Lagrange multipliers.

If I am wrong and you are supposed to use Lagrange multipliers, then you can do that,

and get the same equations as below.

Denote

f(x, y, z) = xy + 12x + z²

Then

∇f = (y+12, x, 2z)

For any point (x, y, z), the vector (x, y, z) connects the origin with the point (x, y, z), and

its length √(x² + y² + z²) is the distance we want to minimize.

It will be locally minimal if the vector (x, y, z) is perpendicular to the surface,

that is, if it is parallel to the gradient ∇f at (x, y, z),

that is, if is there is a multiplier t so that

t(x, y, z) = ∇f

That gives us the three equations

tx = y+12 (1)

ty = x (2)

tz = 2z (3)

Together with the equation of the surface

xy + 12x + z² = 129 (4)

we have four equations and four unknowns.

Case 1: z ≠ 0

Then equation (3) yields t = 2, and

equation (2) yields x = 2y.

Rewriting equation (1)

4y = y + 12

which yields

y = 4

and thus x = 8.

Substituting x and y into equation (4) and simplifying,

z² = 1

Hence z = 1 or z = -1.

Both points (8, 4, 1) and (8, 4, -1) have the same distance 9 from the origin.

So these are both candidates for the closest point.

Case 2. z = 0

Then equations (1), (2), (3), (4) simplify into

tx = y+12

ty = x

xy + 12x = 129

Unfortunately, this set of equations does not have a simple solution.

Instead we can argue that no point (x, y, 0) on the surface can be as close

to the origin as the points (8, 4, 1) and (8, 4, -1) obtained in Case 1.

In case 1 where z = 1 or z = -1, the equation of the surface simplifies to the curve

xy + 12x + 1 = 129, or equivalently

y = f₁(x) = 128/x - 12

In Case 2 where z = 0, the equation of the surface simplifies to the curve

xy + 12x = 129, or equivalently

y = f₂(x) = 129/x - 12

Both f₁ and f₂ represent a hyperbola with a positive and a negative branch.

Lemma:

Assume a given hyperbola y = a/x - b where a > 0 and b > 0.

Then for any point (x, y) on the negative branch of the parabola there is a point

on the positive branch that is closer to the origin.

Proof:

The distance of a point (x, y) to the origin is given by

d(x, y) = √(x² + y²) = √(x² + a²/x² - 2ab/x + b²)

For x < 0, -2ab/x > 0 and therefore d(x, y) > d(-x, y)

This proves the lemma.

Using the Lemma we conclude that for the curve f₂ the closest point to the origin

must lie on the positive branch.

Now we argue that for any such point (x, y), the curve f₁ must have a point even closer.

Notice that for any x > 0, f₁(x) < f₂(x).

Therefore for any x > 0, the point (x, f₁(x)) is closer to the origin that (x, f₂(x)).

The final answer is that the shortest distance is 9 obtained in Case 1.