Coordinate Products with Fixed Sum and Sum of Squares

Why these constraints determine more than a plane

Fix the sum \(x+y+z=s\) and the sum of squares \(x^2+y^2+z^2=q\), where \(s\) and \(q\) are real constants. The first condition describes a plane in three-dimensional space; the second describes a sphere when \(q\geq 0\). Their intersection is usually a circle. We will find the least and greatest possible values of the product \(xyz\) on that intersection.

The constraints and the product are unchanged when the coordinates are permuted. That symmetry will matter: the multiplier equations imply that every extremum in the regular case has at least two equal coordinates. We must still check the degenerate case separately, rather than assume the usual multiplier rule applies everywhere.

First determine when the constraints are feasible

Write \(m=s/3\), the average of the coordinates. Expanding the sum of squared deviations gives

The left-hand side is nonnegative, so feasibility requires \(q\geq s^2/3\). This condition is also sufficient. If it holds, define \(d=\sqrt{(q-s^2/3)/6}\). The triple \((m+d,m+d,m-2d)\) has sum \(3m=s\), and its sum of squares is \(3m^2+6d^2=q\).

For \(q>s^2/3\), the feasible set is closed and bounded: it is the intersection of the plane \(x+y+z=s\) with the sphere of radius \(\sqrt q\). A continuous function on a nonempty closed bounded set attains both its minimum and maximum. Thus \(xyz\) has extrema, and the two-constraint Lagrange multiplier rule can locate them provided the constraint gradients are independent.

The two-multiplier equations

Set \(g(x,y,z)=x+y+z\) and \(h(x,y,z)=x^2+y^2+z^2\). Their gradients are \(\nabla g=(1,1,1)\) and \(\nabla h=(2x,2y,2z)\). These gradients are dependent exactly when \(x=y=z\). On the feasible set, that happens only when \(q=s^2/3\). Therefore, for \(q>s^2/3\), every extremum satisfies \(\nabla f=\lambda\nabla g+\mu\nabla h\) for some real multipliers \(\lambda,\mu\).

Since \(\nabla f=(yz,xz,xy)\), the three coordinate equations are

Subtracting the first two equations gives \(z(y-x)=2\mu(x-y)\), or \((x-y)(z+2\mu)=0\). Subtracting the second and third gives \((y-z)(x+2\mu)=0\). These factored equations retain the cases in which coordinates are equal; dividing by a coordinate difference without checking whether it is zero would not be valid.

Why at least two coordinates must agree

Suppose, for contradiction, that \(x,y,z\) are all distinct at a multiplier candidate. Then \(x-y\ne 0\) and \(y-z\ne 0\), so the two factored equations force \(z=-2\mu\) and \(x=-2\mu\). That gives \(x=z\), contradicting the assumption that all three coordinates are distinct. Hence every multiplier candidate has at least two equal coordinates.

The reduction respects permutation symmetry. If \((a,a,b)\) is feasible, then each permutation of its coordinates is feasible and has the same product \(a^2b\). Thus it is enough to calculate the candidate values for the two possible branches for \(a\), not to list every ordering separately.

Find and compare every candidate value

For a triple \((a,a,b)\), the constraints give \(2a+b=s\), so \(b=s-2a\). Substituting this into the sum-of-squares condition yields

Solving this quadratic gives \(a=s/3\pm d\), where \(d=\sqrt{(3q-s^2)/18}>0\). The corresponding values of \(b=s-2a\) are \(s/3\mp2d\). Put \(m=s/3\). The two candidate products are therefore

Expanding each expression gives

Since \(d>0\), their difference is \(P_{\mathrm{high}}-P_{\mathrm{low}}=4d^3>0\). Both branches give feasible triples, and every global extremum must be among these candidates. Consequently the lower value is the minimum and the higher value is the maximum.

Worked Example: A positive fixed sum

Worked Example: A zero sum

Worked Example: The single-point case

Boundary and exceptional cases

When \(q=s^2/3\), the only feasible triple is \((s/3,s/3,s/3)\), and both extrema equal \((s/3)^3\). The constraint gradients there are \((1,1,1)\) and \((2s/3)(1,1,1)\), so they are dependent. If \(s=0\), the second gradient is the zero vector; the feasible point is \((0,0,0)\), and the product is still determined directly. The multiplier equations for regular points should not be used to justify this single-point case.

If \(q<s^2/3\), there are no feasible triples, so there is no minimum or maximum under the stated constraints. For instance, if \(s=3\) and \(q=2\), then \(q<3=s^2/3\), which rules out feasibility before any multiplier calculation begins.

A common mistake: dividing away the equal-coordinate cases

The subtraction step produces \((x-y)(z+2\mu)=0\), not the unconditional equation \(z+2\mu=0\). Dividing by \(x-y\) assumes \(x\ne y\), and so discards precisely the equal-coordinate cases that include the extrema. In the first worked example, the minimum occurs at \((3,3,0)\), where \(x-y=0\). Dividing by \(x-y\) there is division by zero and cannot be used to derive any conclusion about that candidate.

The correction is to keep the product equal to zero and treat its alternatives: either \(x=y\), or \(z=-2\mu\). Applying this logic to the other subtracted equations shows that three distinct coordinates are impossible, while retaining all cases with equal coordinates. Checking the feasibility condition first and treating the equality case separately completes the argument without losing solutions.

Why the symmetry is useful

Both constraints are symmetric in \(x,y,z\), and so is the product. Permuting a feasible triple preserves its constraint values and objective value. The multiplier equations provide a stronger fact at an extremum: at least two coordinates coincide. Together, these observations reduce an optimization problem on a circle to two candidate product values. Comparing those values—not merely identifying stationary points—is what determines which is the minimum and which is the maximum.