Product Extrema on a Sphere

Why the Product Has Different Extremes on a Sphere

The equation \(x^2+y^2+z^2=r^2\) fixes the distance from the origin, but it does not fix the signs of \(x\), \(y\), or \(z\). Consequently, the product \(xyz\) can be positive, negative, or zero on the same sphere. To find its largest and smallest possible values, we must account for all these possibilities.

Assume first that \(r>0\). The sphere is closed and bounded, hence compact, and \(f(x,y,z)=xyz\) is continuous. Therefore \(f\) attains both a global maximum and a global minimum on the sphere. The Lagrange multiplier method identifies the candidates; comparing their objective values determines the extrema.

Set Up the Multiplier Equations

Write the constraint as \(g(x,y,z)=x^2+y^2+z^2-r^2=0\). Its gradient is \(\nabla g=(2x,2y,2z)\), which is nonzero at every point on the sphere: if all three coordinates were zero, the constraint would give \(r^2=0\), contrary to \(r>0\). Thus the regularity condition for the one-constraint Lagrange theorem holds everywhere on this sphere.

The theorem says that at a constrained extremum of a differentiable objective, when the constraint gradient is nonzero, the objective gradient is a scalar multiple of the constraint gradient. Here \(\nabla f=(yz,xz,xy)\), so for some multiplier \(\lambda\) the equations are:

The equations must be solved with care. Dividing by \(x\), \(y\), or \(z\) before checking whether that coordinate is zero could discard valid candidates. We therefore split the analysis into two cases.

Case One: At Least One Coordinate Is Zero

Suppose \(x=0\). The first multiplier equation becomes \(yz=0\), so either \(y=0\) or \(z=0\). The constraint then forces the remaining coordinate to have magnitude \(r\). This gives the four points \((0,\pm r,0)\) and \((0,0,\pm r)\). Repeating the same reasoning when \(y=0\) or \(z=0\) gives all six axis points:

Each is a multiplier candidate and has objective value \(xyz=0\). For example, at \((r,0,0)\), the first two left sides \(yz\) and \(xz\) are zero, while the first equation requires \(0=2\lambda r\), so \(\lambda=0\); the other equations also hold. The other axis points satisfy the equations by the same check.

Case Two: All Coordinates Are Nonzero

Now suppose \(xyz\ne 0\). To compare the first two equations without dividing by a coordinate or by the multiplier, multiply the first equation by \(x\) and the second by \(y\). Their left sides are both \(xyz\), so their right sides must agree:

In this case \(\lambda\ne 0\), because \(yz\ne 0\) in the first equation. Therefore \(x^2=y^2\). Comparing the second and third equations in the same way gives \(y^2=z^2\). Hence all three squared coordinates are equal. Substituting into the constraint gives:

The coordinates can have different signs, so the product can have either sign. Its magnitude is fixed:

These points really do satisfy the multiplier equations. For any such point, set \(\lambda=xyz/(2x^2)\). Since \(x^2=y^2=z^2\), this is also \(xyz/(2y^2)\) and \(xyz/(2z^2)\), which verifies each equation. There are four sign choices with positive product and four with negative product.

Why Comparing the Candidates Is Enough

The Lagrange equations give necessary conditions for extrema; on their own, they do not say which candidate is a maximum or minimum. The comparison is justified here because the sphere is compact and the objective is continuous, so both global extrema exist. Every global extremum must satisfy the Lagrange equations, as the constraint gradient is nonzero everywhere on the sphere. We found all the candidates: the axis points give \(0\), while the nonzero-coordinate points give the two values \(\pm r^3/(3\sqrt{3})\). Since \(r>0\), the positive value exceeds zero and the negative value is less than zero. Thus the largest and smallest candidate values are the global maximum and minimum, respectively.

Worked Examples

Common Pitfall: Dividing Before Checking the Cases

A tempting shortcut is to divide the equations by \(x\), \(y\), and \(z\) immediately. That step assumes all three coordinates are nonzero and can silently remove valid candidates. For instance, the axis point \((r,0,0)\) satisfies the constraint and the multiplier equations with \(\lambda=0\), but it would be lost in any calculation that divided by \(y\) or \(z\).

The correction is to state the cases first: if a coordinate vanishes, use the equations and constraint to find the axis points; only in the separate case where all coordinates are nonzero may you use nonzero-coordinate algebra. Also, do not label every Lagrange candidate a maximum or minimum. The axis candidates have value zero, and it is the comparison of all candidate values, together with existence of extrema, that identifies the global answers.

Dependence on the Radius

The extremal values scale with the cube of the radius. If the radius is multiplied by a positive factor \(c\), each coordinate at an equal-magnitude candidate is multiplied by \(c\), so their product is multiplied by \(c^3\). The formula \(\pm r^3/(3\sqrt{3})\) records this scaling exactly.

The assumption \(r>0\) matters for the regularity argument. If \(r=0\), the feasible set consists of the single point \((0,0,0)\), so both the maximum and minimum are \(0\); the constraint gradient is zero there and the regular Lagrange condition does not apply.