What Equation Describes a Rectangle with a Given Perimeter?

If you know a rectangle’s perimeter, its length and width are related by a simple equation. The perimeter alone does not usually determine both side lengths: many different rectangles can have the same perimeter.

The standard equation is $P=2l+2w$, where $l$ is the length, $w$ is the width, and $P$ is the perimeter. If you know the area as well, you can combine this equation with the area formula to determine the dimensions.

The perimeter equation

A rectangle has two sides of length $l$ and two sides of width $w$. Adding all four sides gives

$$P=l+w+l+w=2l+2w=2(l+w).$$

If the perimeter $P$ is given, rearrange to express one side in terms of the other:

$$w=\frac{P}{2}-l.$$

This is the equation relating the side lengths. It shows that once you choose a length, the width is fixed if the perimeter must stay the same. For a genuine rectangle, both dimensions must be positive, so $0<l<P/2$ and $0<w<P/2$.

If the area is also given

The area of a rectangle is $A=lw$. Substitute $w=P/2-l$ into that formula:

$$A=l\left(\frac{P}{2}-l\right).$$

This gives the area in terms of one side and the fixed perimeter. If both $P$ and $A$ are known, solve the resulting quadratic equation for $l$:

$$l^2-\frac{P}{2}l+A=0.$$

The two roots, when they are valid, give the rectangle’s side lengths. They may appear in either order because swapping length and width describes the same rectangle.

Worked example: perimeter 40 and area 96

Suppose a rectangle has perimeter $40$ units and area $96$ square units. First use the perimeter equation:

$$40=2(l+w) \quad\Longrightarrow\quad l+w=20 \quad\Longrightarrow\quad w=20-l.$$

Now substitute this into the area equation:

$$96=l(20-l).$$

Rearranging gives

$$l^2-20l+96=0.$$

Factor the quadratic:

$$(l-8)(l-12)=0.$$

Thus $l=8$ or $l=12$. Since the two sides add to $20$, the corresponding widths are $12$ and $8$. The rectangle’s side lengths are therefore $8$ units and $12$ units. Check: $2(8+12)=40$ and $8\cdot12=96$.

What the perimeter alone tells you

For a fixed perimeter, there are many possible rectangles. For example, perimeter $40$ gives $l+w=20$, so side lengths $5$ and $15$, $8$ and $12$, or $10$ and $10$ all satisfy the perimeter condition. Their areas differ, even though their perimeters are equal.

Among rectangles with a fixed perimeter, the square has the greatest area. To see this, write

$$A=l\left(\frac{P}{2}-l\right) = -\left(l-\frac{P}{4}\right)^2+\frac{P^2}{16}.$$

The squared term is never negative, so the area is largest when it is zero: $l=P/4$. Then $w=P/4$ as well, and the maximum area is $P^2/16$.

For perimeter $40$, the maximum area is $40^2/16=100$ square units, achieved by a $10$-by-$10$ square. An area of $96$ is possible, as in the worked example, while an area greater than $100$ is not possible with that perimeter.

Common mistakes

• Using $P=l+w$. That is the sum of two adjacent sides, not the full perimeter. The correct equation is $P=2(l+w)$.
• Assuming one perimeter gives one rectangle. Without another condition, such as the area or one side length, the dimensions are not uniquely determined.
• Mixing units. Perimeter is measured in linear units; area is measured in square units. For instance, a perimeter of $40$ cm and an area of $96$ cm$^2$ use different units for different quantities.
• Accepting impossible roots. A solution must give positive side lengths. If a proposed area exceeds $P^2/16$, no rectangle with that perimeter has that area.