Quadratic Extrema on a Circle

Introduction

A circle is a simple constraint, but optimizing a function on it still requires care. The Lagrange equations may produce points on the coordinate axes, points with both coordinates nonzero, or a whole family of stationary points. Which of these occurs depends on the coefficients of the function.

Here we study a diagonal quadratic function: its squared terms have separate coefficients, and it has no term involving the product \(xy\). The constraint is a circle centered at the origin. We will solve the multiplier equations, check the candidate points, and compare their function values to identify the global extrema.

The Function and the Constraint

Let \(a,b,c\) be real numbers and let \(r>0\). Consider the objective function \(f:\mathbb{R}^2\to\mathbb{R}\) and the constraint \(g(x,y)=0\) defined by

The constant \(c\) shifts every function value by the same amount, so it does not change which points maximize or minimize \(f\). The condition \(r>0\) ensures the constraint is a circle of positive radius. It is closed and bounded, and \(f\) is continuous, so \(f\) attains both an absolute maximum and an absolute minimum on that circle.

At every point on this circle, \(\nabla g(x,y)=(2x,2y)\) is nonzero, because \(r>0\) rules out \(x=y=0\). Thus the regularity condition for the one-constraint Lagrange theorem holds: every constrained extremum satisfies \(\nabla f=\lambda\nabla g\) for some real multiplier \(\lambda\).

Deriving and Solving the Multiplier Equations

The gradients are \(\nabla f=(2ax,2by)\) and \(\nabla g=(2x,2y)\). Therefore the multiplier equations, together with the circle constraint, are

Equivalently, after dividing the first two equations by \(2\),

The product equations must be solved by considering whether each coordinate is zero. The constraint prevents both coordinates from being zero, so every feasible point lies either on an axis or has both coordinates nonzero.

In particular, when \(a\ne b\), there are no mixed stationary points: every constrained extremum must be among the four axis points. When \(a=b\), the objective is \(a(x^2+y^2)+c\), which has the same value at every point of the circle. In that case, all feasible points, including mixed points, are stationary.

Comparing the Candidate Values

At either point \((\pm r,0)\), the objective value is \(ar^2+c\). At either point \((0,\pm r)\), it is \(br^2+c\). Since \(r^2>0\), comparing these values gives the complete result.

Why the Comparison Works

The multiplier calculation identifies the required candidates. A second short argument makes the value comparison especially transparent. On the circle, \(x^2+y^2=r^2\), so we can rewrite the objective as

As a feasible point moves around the circle, \(x^2\) ranges from \(0\) to \(r^2\), and both endpoint values occur. If \(a-b>0\), the displayed expression increases as \(x^2\) increases; its largest value occurs at \(x^2=r^2\), and its smallest at \(x^2=0\). If \(a-b<0\), it decreases as \(x^2\) increases, so the endpoints exchange roles. If \(a-b=0\), the expression is \(br^2\) for every feasible point. This proves the claimed maximum and minimum values and explains why they are attained on the axes unless the coefficients are equal.

This argument also confirms why comparing the multiplier candidates is enough here: the axis points attain the extreme possible values, and the formula covers every point on the circle. The general compactness guarantee ensures extrema exist; the explicit comparison identifies them.

Worked Examples

Common Mistake: Dividing Away an Axis

A frequent error is to divide \((a-\lambda)x=0\) by \(x\) and \((b-\lambda)y=0\) by \(y\), then conclude that \(\lambda=a=b\). That division assumes \(x\ne0\) and \(y\ne0\), so it discards every axis candidate before checking whether an axis is possible.

For example, for \(5x^2+2y^2\) on \(x^2+y^2=9\), dividing by both coordinates would suggest that a stationary point requires \(5=2\), and hence that there are no candidates. But \((3,0)\) is feasible and satisfies the multiplier equations with \(\lambda=5\), while \((0,3)\) is feasible and satisfies them with \(\lambda=2\).

Why This Problem Matters

This example shows how the Lagrange equations translate geometry into algebra. The constraint fixes the total \(x^2+y^2\), while the coefficients \(a\) and \(b\) determine which coordinate direction contributes more to the objective. If one coefficient is larger, the larger squared coordinate gives the larger value; if they are equal, the direction makes no difference.

The case analysis is just as important as solving the equations. Product equations can have solutions because a factor is zero, not only because another factor vanishes. Keeping both possibilities in view prevents lost candidates and makes the comparison reliable.