What are the dimensions of a rectangle with a perimeter of 60 meters that maximizes its area?

To find the dimensions of a rectangle with a perimeter of 60 meters that maximizes its area, we can start with some basic formulas:

The perimeter (P) of a rectangle is given by the formula: P = 2(length + width). In this case, we know the perimeter is 60 meters, so we can write:

60 = 2(length + width)

Dividing both sides by 2 gives us:

30 = length + width

Next, we can express the width (w) in terms of the length (l):

w = 30 – l

The area (A) of a rectangle is given by the formula: A = length × width. Substituting for width, we get:

A = l × (30 – l)

This can be rewritten as:

A = 30l – l²

This equation is a quadratic function in standard form, A = -l² + 30l, which opens downwards (since the coefficient of is negative), indicating that it has a maximum point.

To find the maximum area, we can use the vertex formula for a parabola:

l = -b / 2a

Where a = -1 and b = 30 for our area equation:

l = -30 / (2 × -1) = 15 meters

Now, substituting back to find the width:

w = 30 – l = 30 – 15 = 15 meters

Thus, the dimensions of the rectangle that maximize the area, given a perimeter of 60 meters, are:

Length = 15 meters
Width = 15 meters

Interestingly, this means that the rectangle is actually a square, with equal sides of 15 meters each, providing the largest possible area for the given perimeter.

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