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The area is given by xy and the perimter is given by 2x + 2y, where x is the side length and y is the height. So we have two equations: A = xy, P = 2x + 2y. Because the square and the rectangle have the same perimeter, P will be a constant, say 40. We can simplify this to the equivalent 20 = x + y by dividing both sides by 2. We can then rearrange it to y = 20 - x or vice versa for x and y. By substituting this into the first equation, we get A = x(20 - x) = 20x - x2. By using differentiation, we can find the x value for which A is maximum: A' = 20 - 2x Set A' to 0 (we know this will be a maximum because the x2 coefficient is negative): 0 = 20 - 2x 2x = 20 x = 10. Now we back-substitute this value for x into the equation for y: y = 20 - x = 20 - 10 = 10. Thus, the maximum area occurs when x = 10 and y = 10 i.e. when the shape is a square. This works for any starting value of the perimeter, and means that a square will always have more area than any rectangle with an equal perimeter.

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