The maximum occurs at a square
Among all positive rectangles with a fixed complete perimeter P, the largest area is P²/16. Each side of the maximizing square is P/4. This assumes the entire four-sided boundary is counted and no additional side-length restrictions apply. A practical site with a fixed width, a required opening or one side supplied by an existing wall can create a different problem. State the constraint before applying a familiar maximum-area rule.
Express one side using the perimeter
Since P=2(a+b), the side sum is a+b=P/2. Therefore b=P/2−a and area is A=a(P/2−a). Positive dimensions require 0<a<P/2. For perimeter 40, this becomes b=20−a and A=a(20−a). Choosing side 4 gives the other side 16 and area 64; choosing 8 gives 12 and area 96. The product rises as these unequal sides move toward each other.
Complete the square to prove the limit
Rewrite a(P/2−a) as P²/16−(a−P/4)². The subtracted square cannot be negative, so area cannot exceed P²/16. Equality occurs only when a=P/4, which also makes b=P/4. For P=40, maximum area is 100 at sides 10 and 10. This proof uses no trial-and-error search and explains why moving away from equal sides reduces the area while preserving the same full perimeter.
Measure the loss from unequal sides
At perimeter 40, sides 8 and 12 produce area 96, four below the maximum 100. The deviation of side a from 10 is two, and its square is four, matching the area loss. Sides 6 and 14 have a deviation of four and area loss 16, giving area 84. The loss is a square-unit quantity. A two-unit side change does not mean a two-square-unit area loss because the constraint changes the other side simultaneously.
| Sides | Area | Loss from maximum |
|---|---|---|
| 4 and 16 | 64 | 36 |
| 8 and 12 | 96 | 4 |
| 10 and 10 | 100 | 0 |
All four sides count toward the same fixed boundary.
Check feasibility before recovering a rectangle
If perimeter is 40 and a requested area is 105, no rectangle satisfies both constraints because 105 exceeds 100. Area 96 is feasible and corresponds to sides 8 and 12, up to rotation. The Rectangle Calculator's area-and-perimeter mode validates this boundary. A failed request can also arise from mixed units, such as area in square feet and perimeter in meters, so normalize the quantities before interpreting rejection as a design conclusion.
Do not apply the full-perimeter rule to three sides
Suppose a different problem supplies 40 units of fencing for two widths and one length, with the fourth side already provided by a wall. Its constraint is 2w+l=40 rather than 2w+2l=40. Area is w(40−2w), maximized at w=10 and l=20, giving 200. This is not a counterexample to the square result; it has a different counted boundary. Confirm wall usability and all required sides in a practical application.
Use the calculator to verify candidate shapes
For the original full perimeter of 40, run two-side mode with 4 and 16, 8 and 12, and 10 and 10. All three perimeters must equal 40 while the areas are 64, 96 and 100. The tool verifies dimensions and constraints but does not automatically optimize a site with obstacles, setbacks or installation conditions. Calculate the theoretical target separately, then test whether the actual dimensional restrictions allow that target.
Handle units, scaling and real limits
Doubling a complete perimeter from 40 to 80 doubles the maximizing side from 10 to 20 and quadruples maximum area from 100 to 400. The formula's square explains that scaling. Keep linear units for perimeter and squared units for area, and retain precision before rounding lengths for construction. A rounded side can change the resulting boundary. Material waste, gates, posts and nonrectangular ground are outside the ideal model and need explicitly stated allowances rather than hidden changes to P.
Apply a side-length restriction explicitly
Suppose perimeter remains 40 but a site requires one side to be at most 8. The unrestricted 10-by-10 target is unavailable. Among feasible choices with that bounded side, the best area occurs at 8 by 12, giving 96, because the side can move toward 10 only as far as the restriction permits. Additional minimum lengths or aspect-ratio rules can similarly move the optimum away from the square.
Keep continuous and discrete choices separate. If dimensions must be whole multiples of a construction unit, a theoretical side length may need neighboring feasible candidates rather than a rounded value that exceeds the allowed perimeter. Recompute both area and boundary for each candidate. This is especially useful when the perimeter does not divide into four permitted equal sides. The ideal square remains a bound, while the best buildable choice depends on the stated constraints.