What is the maximum area of a rectangle area with a fixed perimeter of 80 ft?
The sides could be:
20 by 20
15 by 25
10 by 30
Which of these would provide the greatest area?
20 by 20
To find the maximum area of a rectangle with a fixed perimeter, we first need to understand the relationship between the perimeter and area of a rectangle.
The perimeter of a rectangle is calculated by adding the lengths of all four sides. So, for a rectangle with a length of L and width of W, the perimeter would be P = 2L + 2W.
In this case, we have a fixed perimeter of 80 ft, so we can write the equation as follows:
80 = 2L + 2W
To find the maximum area, we need to express the area of the rectangle in terms of a single variable. The area of a rectangle is given by the formula A = L * W.
Now, we can substitute one variable in terms of the other. In this case, let's substitute L in terms of W from the equation for the perimeter:
80 = 2L + 2W
Simplifying, we get:
40 = L + W
L = 40 - W
Substituting L = 40 - W into the formula for the area:
A = L * W
A = (40 - W) * W
A = 40W - W^2
Now, we have the area expressed in terms of a single variable, W. To find the maximum area, we need to find the value of W that maximizes the area. We can do this by finding the vertex of the quadratic equation A = -W^2 + 40W.
The vertex of a quadratic equation in the form A = ax^2 + bx + c is given by the formula x = -b / (2a). In this case, a = -1 and b = 40.
W = -40 / (2*-1)
W = 40 / 2
W = 20
So, the width of the rectangle that maximizes the area is 20 ft. To find the length, we can substitute this value back into the equation L = 40 - W:
L = 40 - 20
L = 20
Therefore, the maximum area of the rectangle with a fixed perimeter of 80 ft is obtained when the width is 20 ft and the length is also 20 ft. The area is calculated as:
A = L * W
A = 20 * 20
A = 400 square feet.