Optimizing A Rectangle With Fixed Perimeter

Why Fixed Perimeter Creates an Optimization Problem

A rectangle’s area depends on two side lengths, while its perimeter records the total length around its boundary. If the perimeter is fixed, making one side longer forces the other side to become shorter. The question is whether some balance of the two lengths produces a greatest possible area, and, if so, which rectangle achieves it.

Let the fixed perimeter be \(P\), measured in units of length. Write the side lengths as \(x\) and \(y\), also measured in units of length. We consider ordinary, nondegenerate rectangles, so both side lengths must be positive. The perimeter condition and area formula are

We will use the perimeter condition to express the area in terms of just one side length. That gives an objective function on its feasible domain. An algebraic identity will then show not only which dimensions give the largest area, but why no other rectangle can tie that maximum.

Build the Area Function and Its Feasible Domain

Assume \(P>0\). Dividing the perimeter equation by \(2\) gives \(x+y=P/2\). Solving for \(y\), then substituting into \(A=xy\), gives

The formula is valid only when the side lengths describe a feasible rectangle. Since \(x>0\) and \(y>0\), the condition \(y=P/2-x>0\) requires \(x<P/2\). Thus the feasible domain is the open interval \(0<x<P/2\). The units are consistent: each term in the area function is a product of lengths, so \(A_P(x)\) is measured in square units.

The endpoints \(x=0\) and \(x=P/2\) are not feasible: at either endpoint one side would have length zero, so the shape would be degenerate rather than a rectangle. They can be useful for understanding how the area behaves near the edge of the domain, but they are not candidate rectangles.

The Algebraic Identity That Locates the Maximum

Expand the area function and complete the square. Every step is an equality, so this rewrites the same objective rather than approximating it:

A square is nonnegative, so \((x-P/4)^2\geq 0\). Subtracting a nonnegative quantity from \(P^2/16\) shows that the area cannot exceed \(P^2/16\). Equality holds exactly when that square is zero, which requires \(x=P/4\). The other side is then \(y=P/2-P/4=P/4\). This value of \(x\) is feasible because \(P>0\) implies \(0<P/4<P/2\).

This argument proves a global maximum directly: it compares the area at every feasible \(x\) with the same upper bound, and it identifies exactly when equality occurs. It also confirms that the maximum is attained inside the feasible domain, even though that domain is open. The nonattainment of the two boundary values does not prevent the interior square from attaining the maximum.

Calculus Check

The completed-square identity is enough to prove the result. Differentiation provides a useful check on the location of the optimum. On the feasible interval,

The only critical point satisfies \(A_P'(x)=0\), so \(P/2-2x=0\) and \(x=P/4\). It lies in the feasible interval. Since \(A_P''(x)=-2<0\), the graph is concave down; in this case, the critical point is a maximum. Substitution into the area function gives \(A_P(P/4)=(P/4)(P/4)=P^2/16\). The algebraic identity supplies the stronger direct comparison with every feasible rectangle and shows that equality occurs only at the square.

Worked Examples

Common Mistake: Treating a Boundary Shape as a Rectangle

A tempting but incorrect approach is to include \(x=0\) or \(x=P/2\) as feasible endpoints because the area function is defined there algebraically. For example, when \(P=40\text{ cm}\), substituting \(x=0\) gives \(A_{40}(0)=0(20-0)=0\text{ cm}^2\). That calculation evaluates the formula, but it does not describe a nondegenerate rectangle: the corresponding side lengths are \(0\text{ cm}\) and \(20\text{ cm}\).

The correction is to distinguish the formula’s endpoint values from the feasible shapes. The rectangle requires \(0<x<P/2\), so the endpoints are excluded. The maximizing value \(x=P/4\), by contrast, lies strictly inside the interval and gives two positive sides. Another common error is to say the maximum occurs when \(x=P/2\), perhaps because that is the largest feasible-looking value of \(x\). In fact, at that boundary the other side would be zero; moreover, the area identity shows that the maximum occurs at the midpoint \(x=P/4\), where the two sides are equal.

Why the Square Result Matters

The result is not just a pattern seen in a few examples. For every positive perimeter \(P\), the same identity writes the area as the fixed upper bound \(P^2/16\) minus a square. Any unequal rectangle has \(x\ne P/4\), so the square term is strictly positive and its area is strictly smaller. This establishes both the optimal dimensions and their uniqueness.

The proof also makes the role of the constraint clear. The two sides cannot vary independently: their sum is fixed at \(P/2\). Moving one side away from half that sum necessarily moves the other in the opposite direction. The algebra quantifies the resulting loss in area as \((x-P/4)^2\). This is why the square is not merely one possible maximizing shape; it is the only one that reaches the maximum.