To find the dimensions of a rectangle with a perimeter of 60 meters that maximizes the area, we start with the formulas for perimeter and area.
The perimeter (P) of a rectangle is given by:
P = 2(l + w)
where l is the length and w is the width. Given that the perimeter is 60m, we can set up the equation:
2(l + w) = 60
This simplifies to:
l + w = 30
Next, we express the area (A) of the rectangle in terms of one variable. The area is given by:
A = l * w
Using the perimeter equation, we can express w as:
w = 30 – l
Now, substitute this expression for w into the area formula:
A = l * (30 – l)
A = 30l – l²
This is a quadratic equation in terms of l. To find the maximum area, we can complete the square or use the vertex formula. The vertex of a quadratic equation of the form ax² + bx + c occurs at l = -b/(2a). In our equation, a = -1 and b = 30. Thus:
l = -30 / (2 * -1) = 15
Now, substitute this value back into the equation for w:
w = 30 – 15 = 15
Therefore, both the length and width are 15 meters, which means the rectangle is a square.
Conclusion: The dimensions of the rectangle that maximizes the area, given a perimeter of 60 meters, are 15m by 15m.