Examples from Constrained extrema

Re-solving the disk problem with multipliers

Optimise

subject to

The gradients are

The multiplier equations are

After dividing the first two equations by $2$,

The unit-circle constraint rules out $(x,y)=(0,0)$, so the coefficient matrix must have determinant zero:

Thus $\lambda=3$ or $\lambda=-1$.

If $\lambda=3$, then $y=x$. The constraint gives $2x^2=1$, so the two same-sign points

give $f=3$.

If $\lambda=-1$, then $y=-x$. The two opposite-sign points

give $f=-1$. This reproduces the boundary part of the closed-disk example, found there by parameterisation.

The closest point on a plane

Find the point on

closest to

The plane has normal vector

The shortest segment from $\mathbf{p}$ to the plane is perpendicular to the plane, so the desired point has the form

Substituting into the plane equation gives

so $-1+6t=3$ and $t=2/3$. Therefore

This nearest point is the orthogonal projection of $\mathbf{p}$ onto the plane. The same point results from minimising

subject to $2x-y+z=3$, but the normal direction makes the geometric solution shorter.

Closest and furthest points on a quartic curve

Find the points on

that are closest to and furthest from the origin. Minimise and maximise

Both expressions depend only on $x^2$ and $y^2$. Set

The problem becomes optimising $s=u+v$ subject to

At an interior constrained extremum in the first quadrant,

The two components on the right must be equal, so

Substituting into the constraint gives

and hence $d^2=u+v=2/3$.

The boundary cases in the first quadrant are $u=0$ or $v=0$. The constraint then gives the other variable as $1$, so $d^2=1$. The feasible set in $(u,v)$ is closed and bounded and these interior and boundary cases exhaust its extrema. Restoring the signs independently, the closest points are

with distance $\sqrt{2/3}$, while the furthest points are

with distance $1$.

The closest point on a parabola

Let

and find the point on $y=x^2$ closest to $\mathbf{x}_0$. A point on the parabola is

The squared distance is

Differentiate:

The stationary values satisfy

The second possibility exists only when $a\ge1/2$. Since $D(x)\to\infty$ as $|x|\to\infty$, compare the stationary candidates:

whereas at $x^2=a-1/2$,

Their difference is

Therefore, if $0<a\le1/2$, the closest point is

If $a>1/2$, the two closest points are

At $a=1/2$ the two non-zero candidates merge with the origin.