Linear Optimization on a Plane-Sphere Intersection

Why This Constraint Pair Is Especially Manageable

A plane cuts a sphere in either no points, one point, or a circle. When the intersection is a circle, optimizing a linear function over it has a clear geometric structure: the objective changes only in directions that lie within the plane. The Lagrange equations capture that structure, while direct feasibility checks ensure that the candidates really belong to both constraints.

We will work in \(\mathbb{R}^3\). Let the objective be \(f(\mathbf{x})=\mathbf{c}\cdot\mathbf{x}\), where \(\mathbf{x}=(x,y,z)\), and write the plane and sphere as \(\mathbf{a}\cdot\mathbf{x}=b\) and \(\mathbf{x}\cdot\mathbf{x}=R^2\). Here \(\mathbf{a}\ne\mathbf{0}\) and \(R>0\).

Definition: The feasible set is the collection of points satisfying both constraints: \(\{\mathbf{x}\in\mathbb{R}^3:\mathbf{a}\cdot\mathbf{x}=b,\ \mathbf{x}\cdot\mathbf{x}=R^2\}\). A feasible candidate must satisfy both equations; satisfying the multiplier equations alone is not a feasibility check.

Check Feasibility Before Optimizing

The point of the plane closest to the origin is \(\mathbf{p}=\dfrac{b}{\|\mathbf{a}\|^2}\mathbf{a}\). Every point on the plane can be written as \(\mathbf{x}=\mathbf{p}+\mathbf{u}\), where \(\mathbf{a}\cdot\mathbf{u}=0\). Since \(\mathbf{p}\) is parallel to \(\mathbf{a}\), it is perpendicular to \(\mathbf{u}\). Thus \(\|\mathbf{x}\|^2=\|\mathbf{p}\|^2+\|\mathbf{u}\|^2\), and \(\|\mathbf{p}\|=|b|/\|\mathbf{a}\|\).

\begin{array}{ll} |b|>R\|\mathbf{a}\| &: \text{the plane and sphere do not intersect},\\ |b|=R\|\mathbf{a}\| &: \text{they meet at one point},\\ |b|<R\|\mathbf{a}\| &: \text{they meet in a circle of radius } \rho=\sqrt{R^2-\dfrac{b^2}{\|\mathbf{a}\|^2}}. \end{array}

The main method below concerns the circle case. It is compact, so a continuous objective attains both a maximum and a minimum there. The one-point case can be evaluated directly; the empty case has no feasible optimization problem.

The Two-Constraint Lagrange System

Write the constraints as \(g(\mathbf{x})=\mathbf{a}\cdot\mathbf{x}-b=0\) and \(h(\mathbf{x})=\mathbf{x}\cdot\mathbf{x}-R^2=0\). Their gradients are \(\nabla g=\mathbf{a}\) and \(\nabla h=2\mathbf{x}\). At a point on a genuine intersection circle these gradients are independent, so the multiple-constraint Lagrange condition applies: at a constrained extremum, the objective gradient is a linear combination of the constraint gradients.

Theorem: On a plane-sphere intersection circle, any constrained extremum of \(f(\mathbf{x})=\mathbf{c}\cdot\mathbf{x}\) satisfies \(\mathbf{c}=\lambda\mathbf{a}+2\mu\mathbf{x}\), together with both constraints. The scalar multipliers are \(\lambda\) and \(\mu\).

\begin{cases} \mathbf{c}=\lambda\mathbf{a}+2\mu\mathbf{x},\\ \mathbf{a}\cdot\mathbf{x}=b,\\ \mathbf{x}\cdot\mathbf{x}=R^2. \end{cases}

In coordinates, the first vector equation means three scalar equations, one for each coordinate. The multiplier condition gives necessary candidates, not automatically a maximum or minimum. We must solve the constraints, check feasibility, and compare objective values—or prove why the candidates are global extrema.

Projecting the Objective into the Plane

Define the component of \(\mathbf{c}\) parallel to the plane by removing its component in the normal direction \(\mathbf{a}\):

\mathbf{q} = \mathbf{c} - \frac{\mathbf{a}\cdot\mathbf{c}}{\|\mathbf{a}\|^2}\mathbf{a}.

Then \(\mathbf{a}\cdot\mathbf{q}=0\), so \(\mathbf{q}\) is parallel to the plane. For \(\mathbf{x}=\mathbf{p}+\mathbf{u}\), the plane constraint gives \(\mathbf{a}\cdot\mathbf{u}=0\), and the sphere constraint gives \(\|\mathbf{u}\|=\rho\). The objective becomes \(f(\mathbf{x})=\mathbf{c}\cdot\mathbf{p}+\mathbf{q}\cdot\mathbf{u}\). The first term is fixed; only the second varies over the circle.

Key result: If \(|b|<R\|\mathbf{a}\|\) and \(\mathbf{q}\ne\mathbf{0}\), the maximum and minimum occur at \(\mathbf{x}_{\max}=\mathbf{p}+\rho\dfrac{\mathbf{q}}{\|\mathbf{q}\|}\) and \(\mathbf{x}_{\min}=\mathbf{p}-\rho\dfrac{\mathbf{q}}{\|\mathbf{q}\|}\). Their objective values are \(\mathbf{c}\cdot\mathbf{p}+\rho\|\mathbf{q}\|\) and \(\mathbf{c}\cdot\mathbf{p}-\rho\|\mathbf{q}\|\), respectively.

Why the Candidates Are the Global Extrema

Here is both a justification of the result and a way to see why the multiplier equations lead to these points. At any feasible point, \(\|\mathbf{u}\|=\rho\), and the Cauchy–Schwarz inequality gives \(-\|\mathbf{q}\|\rho\le\mathbf{q}\cdot\mathbf{u}\le\|\mathbf{q}\|\rho\). If \(\mathbf{q}\ne\mathbf{0}\), equality on the right requires \(\mathbf{u}\) to point in the direction of \(\mathbf{q}\); equality on the left requires it to point in the opposite direction. Both directions lie in the plane because \(\mathbf{q}\) does. They therefore give feasible points on the circle and attain the bounds. This proves the maximum and minimum formulas, not merely that they are local candidates.

The multiplier equations recover the same two choices. Substitute \(\mathbf{x}=\mathbf{p}+\mathbf{u}\) into \(\mathbf{c}=\lambda\mathbf{a}+2\mu\mathbf{x}\), then project onto directions perpendicular to \(\mathbf{a}\). The \(\lambda\mathbf{a}\) and \(2\mu\mathbf{p}\) terms have no component in those directions, leaving \(\mathbf{q}=2\mu\mathbf{u}\). Since \(\|\mathbf{u}\|=\rho\), this gives \(|2\mu|=\|\mathbf{q}\|/\rho\), and hence the two opposite choices of \(\mathbf{u}\). Taking the dot product of the multiplier equation with \(\mathbf{a}\) then yields \(\lambda=(\mathbf{a}\cdot\mathbf{c}-2\mu b)/\|\mathbf{a}\|^2\).

Worked Examples

Worked Example: A Circle Centered at the Origin

Maximize and minimize \(f(x,y,z)=x+2y+2z\), subject to \(x+y+z=0\) and \(x^2+y^2+z^2=6\).

Here \(\mathbf{a}=(1,1,1)\), \(b=0\), \(R=\sqrt{6}\), and \(\mathbf{c}=(1,2,2)\). The plane-sphere intersection is a circle because \(0<\sqrt{6}\sqrt{3}\). Its center is \(\mathbf{p}=(0,0,0)\), and its radius is \(\rho=\sqrt{6}\). The squared norm of \(\mathbf{a}\) is \(3\), and \(\mathbf{a}\cdot\mathbf{c}=5\), so

\mathbf{q} =(1,2,2)-\frac{5}{3}(1,1,1) =\left(-\frac{2}{3},\frac{1}{3},\frac{1}{3}\right), \qquad \|\mathbf{q}\|=\frac{\sqrt{6}}{3}.

Thus \(\rho/\|\mathbf{q}\|=3\). The two candidates are \(\mathbf{x}_{\max}=(-2,1,1)\) and \(\mathbf{x}_{\min}=(2,-1,-1)\). For the first, the plane constraint gives \(-2+1+1=0\), and the sphere constraint gives \(4+1+1=6\). Its objective value is \(-2+2+2=2\). For the second, the plane sum is \(2-1-1=0\), the squared norm is \(4+1+1=6\), and the objective value is \(2-2-2=-2\).

The multiplier values also check. For \(\mathbf{x}_{\max}\), \(2\mu=\|\mathbf{q}\|/\rho=1/3\), so \(\mu=1/6\); since \(b=0\), \(\lambda=5/3\). Direct substitution gives \((5/3)(1,1,1)+(1/3)(-2,1,1)=(1,2,2)=\mathbf{c}\). For \(\mathbf{x}_{\min}\), \(\mu=-1/6\), \(\lambda=5/3\), and \((5/3)(1,1,1)-(1/3)(2,-1,-1)=(1,2,2)\). The verified global maximum is \(2\), and the global minimum is \(-2\).

Worked Example: A Shifted Intersection Circle

Optimize \(f(x,y,z)=x\) subject to \(x+y+z=1\) and \(x^2+y^2+z^2=3\).

Here \(\mathbf{a}=(1,1,1)\), \(b=1\), \(R=\sqrt{3}\), and \(\mathbf{c}=(1,0,0)\). Since \(\|\mathbf{a}\|^2=3\), the center is \(\mathbf{p}=(1/3,1/3,1/3)\). The circle radius is \(\rho=\sqrt{3-1/3}=\sqrt{8/3}\). Also, \(\mathbf{a}\cdot\mathbf{c}=1\), so

\mathbf{q}=(1,0,0)-\frac{1}{3}(1,1,1) =\left(\frac{2}{3},-\frac{1}{3},-\frac{1}{3}\right), \qquad \|\mathbf{q}\|=\frac{\sqrt{6}}{3}, \qquad \frac{\rho}{\|\mathbf{q}\|}=2.

The maximum candidate is \(\mathbf{x}_{\max}=(1/3,1/3,1/3)+2\mathbf{q}=(5/3,-1/3,-1/3)\). Its coordinates sum to \(1\), its squared norm is \(25/9+1/9+1/9=3\), and \(f=5/3\). The minimum candidate is \(\mathbf{x}_{\min}=(1/3,1/3,1/3)-2\mathbf{q}=(-1,1,1)\). Its coordinates sum to \(1\), its squared norm is \(1+1+1=3\), and \(f=-1\).

For the maximum, \(\mu=\|\mathbf{q}\|/(2\rho)=1/4\) and \(\lambda=(1-2\mu b)/3=1/6\). Then \(\lambda\mathbf{a}+2\mu\mathbf{x}_{\max}=(1,0,0)\). For the minimum, \(\mu=-1/4\) and \(\lambda=(1-2\mu b)/3=1/2\); then \(\lambda\mathbf{a}+2\mu\mathbf{x}_{\min}=(1,0,0)\). The constraints and stationarity equation check for both candidates, and the global bounds show that \(5/3\) and \(-1\) are the maximum and minimum.

Worked Example: An Objective Constant on the Circle

Optimize \(f(x,y,z)=3x\) subject to \(x=0\) and \(x^2+y^2+z^2=4\).

The constraints describe the circle \(x=0,\ y^2+z^2=4\), which is nonempty. Every feasible point has \(f=3(0)=0\), so the maximum and minimum are both \(0\), attained at every point of the circle. In vector notation, \(\mathbf{a}=(1,0,0)\), \(\mathbf{c}=(3,0,0)=3\mathbf{a}\), and therefore \(\mathbf{q}=\mathbf{0}\). The multiplier equation holds at every feasible point with \(\lambda=3\) and \(\mu=0\): \(\lambda\mathbf{a}+2\mu\mathbf{x}=(3,0,0)=\mathbf{c}\). The objective is constant because it points normal to the plane, not along the circle.

Common Pitfall: Using the Full Objective Direction

A tempting shortcut is to move from the circle center in the direction of \(\mathbf{c}\). That direction may point out of the plane, so the resulting point need not be feasible. For the shifted-circle example, the center is \((1/3,1/3,1/3)\), and \(\mathbf{c}=(1,0,0)\). Moving a distance \(\rho=\sqrt{8/3}\) in the unit direction of \(\mathbf{c}\) gives \((1/3+\sqrt{8/3},1/3,1/3)\). Its coordinate sum is \(1+\sqrt{8/3}\), not \(1\), so it is not on the plane. The correction is to use \(\mathbf{q}\), the component of the objective parallel to the plane, and move in the direction \(\mathbf{q}/\|\mathbf{q}\|\).

There are two further cases to diagnose. If \(\mathbf{q}=\mathbf{0}\), the objective is constant on the intersection circle; do not report two distinct extremizers. If \(|b|=R\|\mathbf{a}\|\), the plane is tangent to the sphere and the feasible set is one point. Evaluate the objective there directly rather than treating the point as a regular intersection circle.

Check Your Understanding

Use the plane, sphere, and objective equations to answer each question. For multiplier candidates, check both constraints.

For the plane \(x+y=1\) and the sphere \(x^2+y^2+z^2=4\), find the circle center and radius.
For \(\mathbf{a}=(1,1,0)\) and \(\mathbf{c}=(1,0,1)\), compute the projected vector \(\mathbf{q}\).
Why can moving in the direction of \(\mathbf{c}\) fail to give a feasible candidate?
What does \(\mathbf{q}=\mathbf{0}\) imply about a linear objective on the intersection circle?
State the feasibility condition for the plane \(\mathbf{a}\cdot\mathbf{x}=b\) to intersect the sphere \(\|\mathbf{x}\|=R\).