To find the dimensions of a rectangle that maximizes the area while having a perimeter of 100 meters, we can start with some basic formulas.
The perimeter (P) of a rectangle is given by the formula:
P = 2(l + w)
Where l is the length and w is the width of the rectangle. Given that the perimeter is 100 m, we can set up the equation:
100 = 2(l + w)
Simplifying this, we find:
l + w = 50
Now, to express the area (A) of the rectangle, we use the formula:
A = l × w
To express area in terms of one variable, we can substitute w from the perimeter equation:
w = 50 – l
Now substituting for w in the area formula gives:
A = l × (50 – l) = 50l – l²
This is a quadratic equation that opens downward (as the coefficient of l² is negative). The maximum area occurs at the vertex of this parabola. The formula for the l coordinate of the vertex of a quadratic equation Ax² + Bx + C is given by:
l = -B / (2A)
In our case, A = -1 and B = 50, so:
l = -50 / (2 × -1) = 25
Thus, the length l should be 25 meters. Using the perimeter equation, we can find w:
w = 50 – l = 50 – 25 = 25
So the dimensions of the rectangle that maximizes the area, while keeping the perimeter at 100 m, are:
Length: 25 m
Width: 25 m
This means the rectangle is actually a square. The maximum area for this rectangle is:
A = 25 × 25 = 625 m²