Maximizing An Inscribed Rectangle In A Circle

From Inscribed Geometry to an Area Function

A rectangle inscribed in a circle has all four vertices on the circle. Its dimensions cannot vary independently: making it wider changes how tall it can be while still fitting inside the same circle. Optimization turns this geometric restriction into a one-variable problem.

We will consider a nondegenerate rectangle whose sides are parallel to the coordinate axes, inscribed in a circle of radius \(R>0\). The circle may have any center. We will show that the unique largest-area rectangle is a square, and find its dimensions and area.

Why the Rectangle Is Centered at the Circle’s Center

First place the circle’s center at \((a,b)\). The rectangle’s horizontal coordinates are \(u\) and \(v\), with \(u<v\), and its vertical coordinates are \(s\) and \(t\), with \(s<t\). Its four vertices are \((u,s)\), \((u,t)\), \((v,s)\), and \((v,t)\).

The circle equation is \((X-a)^2+(Y-b)^2=R^2\). Since all four vertices lie on the circle, compare the equations for \((u,s)\) and \((v,s)\). Subtracting gives

Because \(u\ne v\), it follows that \(u+v=2a\). Repeating the comparison for \((u,s)\) and \((u,t)\) gives \(s+t=2b\). Thus the rectangle’s center is \((a,b)\), the circle’s center.

Let \(x=(v-u)/2\) be the rectangle’s half-width and \(y=(t-s)/2\) its half-height. Both are positive. The top-right vertex is \((a+x,b+y)\), so its membership in the circle gives the constraint \(x^2+y^2=R^2\). The full width is \(2x\), and the full height is \(2y\).

Build the One-Variable Objective

From \(x^2+y^2=R^2\), the positive half-height corresponding to a chosen half-width \(x\) is \(y=\sqrt{R^2-x^2}\). A nondegenerate rectangle has \(0<x<R\). For checking the limiting cases, extend the area function to the closed interval \(0\le x\le R\); at either endpoint the area is zero, corresponding to a degenerate rectangle.

This function is continuous on the closed interval, so it attains a maximum there. On the interior, differentiate using the product rule and the chain rule:

For \(0<x<R\), the denominator is positive. Therefore an interior critical point occurs exactly when \(R^2-2x^2=0\). Since \(x>0\), the solution is \(x=R/\sqrt{2}\). The constraint then gives \(y=R/\sqrt{2}\) as well.

Why This Critical Point Is the Global Maximum

A critical point alone does not establish a global maximum. We must also account for the endpoints and determine how the area changes across the feasible interval. The derivative provides that information: its sign is the sign of \(R^2-2x^2\).

The function therefore increases to the critical point and decreases afterward. That point is the unique global maximum on the full feasible interval. Substituting \(x=y=R/\sqrt{2}\), the full side lengths are \(2x=2y=\sqrt{2}R\), and the maximum area is \(4xy=2R^2\).

A Geometric Check on the Result

At the maximum, the two half-dimensions are equal. The rectangle is consequently a square, and the circle constraint confirms its diagonal: if each side is \(\sqrt{2}R\), the square’s diagonal has length

This agrees with the fact that the square’s opposite vertices lie at opposite ends of a diameter. The area of the largest rectangle is \(2/\pi\) times the area \(\pi R^2\) of the circle. That ratio is the same for every radius: changing the circle’s size scales all lengths by the same factor and areas by its square.

Worked Examples

Common Pitfall: Confusing Half-Dimensions with Side Lengths

The variables \(x\) and \(y\) in the constraint \(x^2+y^2=R^2\) measure from the circle’s center to one corner. They are half the rectangle’s width and height. Forgetting that distinction loses a factor of four in the area.

For example, with \(R=5\), the maximizing half-dimensions are \(x=y=5/\sqrt{2}\). A mistaken calculation might call these the full side lengths and report an area of \((5/\sqrt{2})^2=25/2\) square units. But each full side is twice as long, so the correct area is

A reliable check is to calculate the rectangle’s dimensions from the center-based variables before calculating area: width \(=2x\), height \(=2y\), and area \(=(2x)(2y)\).

When This Optimization Method Is Useful

The circle equation is the constraint, and the rectangle’s area is the objective. Expressing one half-dimension in terms of the other reduces the problem to a single variable; checking the critical point together with the boundary completes the global comparison. This same structure appears whenever a geometric restriction links dimensions that would otherwise vary independently.

The answer also captures a general geometric tradeoff. A rectangle can use more of the available radius in one direction only by using less in the perpendicular direction. At the maximum, neither direction is favored: the half-width and half-height are equal. The result is not just a plausible symmetric guess; the derivative and endpoint checks establish that it is the unique largest-area choice.