Eighty metres of fencing are available to enclose a rectangular play area. Determine the dimensions that will enclosed the maximum area. What is the maximum area?
2(x+y)=80, so y = 40-x
area = xy = x(40-x) = 40x-x^2
That's just a parabola. Find the vertex and you will have the maximum area.
A little investigation that the rectangular shape of maximum area for a given perimeter is always a square.
To determine the dimensions that will enclose the maximum area, we can use the formula for the perimeter of a rectangle:
Perimeter = 2 * (Length + Width)
Since we are given that the perimeter is 80 meters, we can set up the equation:
80 = 2 * (Length + Width)
Simplifying this equation, we have:
40 = Length + Width
To find the maximum area, we can use the formula for the area of a rectangle:
Area = Length * Width
We want to maximize the area, so we can express the length in terms of the width. Rearranging the equation from above, we get:
Length = 40 - Width
Substituting this into the area equation, we have:
Area = (40 - Width) * Width
To find the maximum area, we can take the derivative of the area equation with respect to the width and set it to zero. Let's do that:
d(Area)/d(Width) = 0
d((40 - Width) * Width)/d(Width) = 0
40 - 2 * Width = 0
2 * Width = 40
Width = 40 / 2 = 20
Now, we can substitute the width back into the length equation to find the length:
Length = 40 - Width = 40 - 20 = 20
Therefore, the dimensions that will enclose the maximum area are a length of 20 meters and a width of 20 meters. The maximum area can be calculated by substituting these dimensions into the area equation:
Area = Length * Width = 20 * 20 = 400 square meters
Therefore, the maximum area is 400 square meters.
To determine the dimensions that will enclose the maximum area using 80 meters of fencing, we can follow these steps:
1. Let's assume the length of the rectangular play area is "L" meters and the width is "W" meters.
2. The total amount of fencing required will be equal to the perimeter of the rectangular play area, which is given as 80 meters. So, we can set up the equation: 2L + 2W = 80.
3. We can simplify the equation by dividing both sides by 2, which gives us: L + W = 40.
4. To express one variable in terms of the other, we'll rearrange the equation to solve for either L or W. Let's solve for L: L = 40 - W.
5. The area of a rectangle is given by the formula: A = L * W.
6. Substituting the value of L from step 4 into the area formula, we get: A = (40 - W) * W.
7. Now, we have the area of the rectangle as a function of only one variable, W. We can express this function as A(W) = 40W - W^2.
8. To find the maximum area, we need to find the value of W that maximizes the function A(W).
9. Since the function is a quadratic equation with a negative coefficient for the squared term (-1), we know the graph will look like a downward-opening parabola. The maximum value will occur at the vertex of this parabola.
10. The x-coordinate of the vertex of a quadratic function in the form A(x) = ax^2 + bx + c is given by the formula: x = -b / (2a).
11. In our case, a = -1 and b = 40. Substituting these values into the formula, we find: W = -40 / (2 * (-1)) = 20.
12. We found that the width of the rectangle that maximizes the area is 20 meters. To find the corresponding length, we can substitute this value back into our equation from step 4: L = 40 - W = 40 - 20 = 20.
13. Therefore, the dimensions that will enclose the maximum area are: length = 20 meters and width = 20 meters.
14. Finally, we can find the maximum area by substituting the width and length values into the area formula: A = L * W = 20 * 20 = 400 square meters.
So, the maximum area that can be enclosed with 80 meters of fencing is 400 square meters.