if the perimeter of a rectangle is 50m what is the largest area possible
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To find the largest possible area of a rectangle with a given perimeter, we can use the concept of optimization. The formula for the perimeter (P) of a rectangle is given by:
P = 2l + 2w
where l represents the length and w represents the width of the rectangle.
In this case, we are given that the perimeter of the rectangle is 50 meters. So, we can rewrite the formula as:
50 = 2l + 2w
Let's rearrange this equation to express w in terms of l:
50 - 2l = 2w
25 - l = w
The area (A) of a rectangle is given by the formula:
A = lw
We want to maximize the area, A. So, we substitute w = 25 - l into the area formula:
A = l(25 - l)
Now, we can expand the equation to get a quadratic equation in the form of Ax^2 + Bx + C:
A = -l^2 + 25l
To find the largest area, we need to find the vertex of the quadratic equation. The x-coordinate of the vertex gives us the value of l that maximizes the area.
The x-coordinate of the vertex can be found using the formula:
x = -B / (2A)
In this case, A = -1 and B = 25. Substituting these values into the formula:
l = -25 / (2(-1))
l = 25 / 2
l = 12.5
Now, we can substitute this value back into the expression for w:
w = 25 - l
w = 25 - 12.5
w = 12.5
Therefore, the length (l) and width (w) of the rectangle that will result in the largest possible area with a perimeter of 50 meters are 12.5 meters each.
To find the corresponding maximum area (A), substitute the values of l and w into the area formula:
A = l * w
A = 12.5 * 12.5
A = 156.25 square meters
Hence, the largest possible area of the rectangle is 156.25 square meters.