Why a Disk Requires an Interior and a Boundary Check
A polynomial can attain its largest or smallest value on a closed disk either at an interior point or on the circle bounding the disk. Searching only the circle can therefore miss an extremum. Searching only for points where both partial derivatives vanish can miss extrema that occur at the boundary.
The method in this workshop treats these two possibilities separately. First, find the interior critical points. Then restrict the polynomial to the circle, find the stationary points of that one-variable function, and compare all the resulting function values.
A convenient parametrization of this circle is \(x=a+R\cos t,\ y=b+R\sin t\), where \(0\le t\le 2\pi\). Given a polynomial \(f(x,y)\), its boundary restriction is the one-variable function \(g(t)=f(a+R\cos t,b+R\sin t)\).
The Candidate-Set Principle
A polynomial is continuous, and a closed disk is compact, so the extreme value theorem guarantees that \(f\) attains an absolute maximum and an absolute minimum on the disk. The following argument explains where those values must be found.
Here \(f_x\) and \(f_y\) denote the partial derivatives. The theorem gives necessary candidate conditions, not a guarantee that every candidate is an extremum. Once the candidates are found, their function values must be compared.
Solve \(f_x(x,y)=0\) and \(f_y(x,y)=0\), retaining solutions strictly inside the disk.
Substitute \(x=a+R\cos t\) and \(y=b+R\sin t\) into \(f\) to obtain \(g(t)\).
Solve \(g'(t)=0\) for \(0\le t<2\pi\), or identify when the derivative vanishes for every \(t\).
Evaluate \(f\) at the interior candidates and the corresponding boundary points. The least and greatest values are the absolute minimum and maximum.
Why These Conditions Find the Extrema
Suppose an absolute extremum occurs at a point strictly inside the disk. There is a small open circle around that point that stays inside the disk. In particular, we can vary \(x\) a little while fixing \(y\), and we can vary \(y\) a little while fixing \(x\). Each resulting one-variable function has a local extremum. By the one-variable necessary condition for a differentiable function to have a local extremum at an interior point, its derivative there is zero. Those derivatives are \(f_x\) and \(f_y\), so both must vanish.
Now suppose an absolute extremum occurs on the boundary. Every point of the circle is represented by the parametrization, so the corresponding parameter value is a local extremum of \(g(t)\). The parameter can be considered on the real line, with the circle repeating every \(2\pi\); there is no missing endpoint of the circle at \(t=0\). The one-variable necessary condition gives \(g'(t)=0\). If \(g\) is constant, this condition holds at every boundary point.
These arguments show that at least one absolute maximum and one absolute minimum appear in the candidate search. They do not say that every candidate is an extremum, or that there are only finitely many boundary candidates. Comparing the values is what determines the answer.
Worked Example: A Linear Term Creates an Interior Minimum
Worked Example: A Linear Term Creates an Interior Minimum
Find the absolute extrema of \(f(x,y)=x^2+y^2+x\) on the disk \(x^2+y^2\le 4\).
Interior candidates. Differentiate: \(f_x=2x+1\) and \(f_y=2y\). Setting both equal to zero gives \(2x+1=0\), so \(x=-\frac12\), and \(2y=0\), so \(y=0\). This point is inside the disk because \((-\frac12)^2+0^2=\frac14<4\). Its value is \(f(-\frac12,0)=\frac14-\frac12=-\frac14\).
Boundary candidates. Parametrize the circle of radius \(2\) by \(x=2\cos t,\ y=2\sin t\). Substitution gives \(g(t)=(2\cos t)^2+(2\sin t)^2+2\cos t=4+2\cos t\). Thus \(g'(t)=-2\sin t\), which is zero when \(t=0\) or \(t=\pi\) on \(0\le t<2\pi\). At \(t=0\), the point is \((2,0)\), and \(f(2,0)=4+2=6\). At \(t=\pi\), the point is \((-2,0)\), and \(f(-2,0)=4-2=2\).
Compare and check. The candidate values are \(-\frac14,2,6\). Therefore the absolute minimum is \(-\frac14\) at \((-\frac12,0)\), and the absolute maximum is \(6\) at \((2,0)\). Substitution confirms the minimum point satisfies the disk constraint, while \((2,0)\) satisfies \(2^2+0^2=4\) and lies on the boundary.
Worked Example: A Mixed Term on the Unit Disk
Worked Example: A Mixed Term on the Unit Disk
Find the absolute extrema of \(f(x,y)=x^2+y^2+xy\) on the unit disk \(x^2+y^2\le 1\).
Interior candidates. The partial derivatives are \(f_x=2x+y\) and \(f_y=x+2y\). From \(2x+y=0\), we have \(y=-2x\). Substitute this into \(x+2y=0\): \(x+2(-2x)=-3x=0\), so \(x=0\) and \(y=0\). The point lies inside the disk, and \(f(0,0)=0\).
Boundary candidates. On the unit circle, \(x=\cos t\) and \(y=\sin t\). Hence \(g(t)=\cos^2t+\sin^2t+\cos t\sin t=1+\cos t\sin t\). Differentiating gives \(g'(t)=\cos^2t-\sin^2t=\cos(2t)\). The stationary parameters in \(0\le t<2\pi\) satisfy \(2t=\frac{\pi}{2}+k\pi\), giving \(t=\frac{\pi}{4},\frac{3\pi}{4},\frac{5\pi}{4},\frac{7\pi}{4}\). At these parameters, \(\cos t\sin t\) alternates between \(\frac12\) and \(-\frac12\), so the boundary values are \(\frac32\) and \(\frac12\).
Compare and check. The candidates are the interior value \(0\) and boundary values \(\frac12,\frac32\). The absolute minimum is \(0\) at \((0,0)\); the absolute maximum is \(\frac32\), attained where \(\cos t\sin t=\frac12\), for example at \((\frac{1}{\sqrt2},\frac{1}{\sqrt2})\). Direct substitution gives \(f(\frac{1}{\sqrt2},\frac{1}{\sqrt2})=\frac12+\frac12+\frac12=\frac32\), and its squared distance from the origin is \(\frac12+\frac12=1\), as required.
Worked Example: A Cubic with Several Boundary Extrema
Worked Example: A Cubic with Several Boundary Extrema
Find the absolute extrema of \(f(x,y)=x^3-3xy^2\) on the unit disk.
Interior candidates. The partial derivatives are \(f_x=3x^2-3y^2\) and \(f_y=-6xy\). For both to vanish, we need \(x^2-y^2=0\) and \(xy=0\). If \(x=0\), then \(x^2-y^2=0\) forces \(y=0\). If \(y=0\), that same equation forces \(x=0\). Thus the only interior critical point is \((0,0)\), with value \(f(0,0)=0\).
Boundary candidates. Substitute \(x=\cos t,\ y=\sin t\): \(g(t)=\cos^3t-3\cos t\sin^2t=\cos(3t)\). Therefore \(g'(t)=-3\sin(3t)\), which vanishes when \(3t=k\pi\). For \(0\le t<2\pi\), this gives \(t=k\pi/3\) for \(k=0,1,2,3,4,5\). The corresponding values are \(g(k\pi/3)=\cos(k\pi)=(-1)^k\). Thus the boundary values are \(1\) and \(-1\).
Compare and check. Comparing \(-1,0,1\), the absolute minimum is \(-1\) and the absolute maximum is \(1\). For example, at \(t=0\), the boundary point is \((1,0)\) and \(f(1,0)=1\). At \(t=\pi/3\), the point is \((\frac12,\frac{\sqrt3}{2})\), and \(f(\frac12,\frac{\sqrt3}{2})=\frac18-3(\frac12)(\frac34)=\frac18-\frac98=-1\). Both points lie on the unit circle, and these evaluations check the boundary values.
A Common Pitfall: Checking Only the Circle
In the first example, the smallest value on the boundary was \(2\), attained at \((-2,0)\). If we incorrectly checked only the boundary, we would report \(2\) as the absolute minimum. But the interior point \((-\frac12,0)\) has value \(-\frac14\), which is smaller. The correct minimum is therefore \(-\frac14\).
The reverse omission is also possible: solving only \(f_x=f_y=0\) does not identify boundary extrema that are not interior critical points. The candidate-set method avoids both errors by treating the interior and circle separately, then comparing their values.
Check Your Understanding
For each question, identify the candidate conditions before deciding how to compare values.
- For a polynomial on a closed disk, what two types of points must be checked for absolute extrema?
- What boundary function results from restricting \(f(x,y)\) to a disk centered at \((a,b)\) with radius \(R\)?
- Why must both partial derivatives vanish at an interior extremum?
- For \(f(x,y)=x^2+y^2\) on the unit disk, what is the boundary restriction, and why does its derivative provide no finite list of boundary points?
- In the first worked example, what incorrect minimum would result from checking only the circle, and what interior candidate corrects it?