24 m of fencing are available to enclose a play area. what would the maximum area be if the fencing only went on 3 sides because the wall is used as the 4th side.
Please show you work
l + 2w = perimeter
l + 2w = 24
l = 24-2w
Area = l times w
(24-2w)w = Area.
24w -2w^2 = Area.
If you are doing this using calculus, you would take the derivative of the area and set it equal to zero to find w.
let the side parallel to the wall by y m
let the other two sides be x m each
so we have y + 2x = 24 or y = 24-2x
area = xy
= x(24-2x
= -2x^2 + 24x
let's complete the square to find the vertex of this parabola
= -2(x^2 - 12x + 36 - 36)
= -2( (x-6)^2 - 36)
= -2(x-6)^2 + 72
the vertex is (6,72)
So the maximumum area is 72 ^2 , when x = 6 and y = 12
or
by calculus
d(area) = -4x + 24 = 0 for a max area
x = 6
then y = 24 - 12 = 12
max area = 6(12) = 72
To find the maximum area using only three sides of fencing while the fourth side is a wall, we need to determine the dimensions that would maximize the enclosed area.
Let's assume the play area is a rectangle. Let 'x' represent the width of the rectangle and 'y' represent the length.
In this case, the available fencing would be used for two sides of width and one side of length. There are two widths, so the total length of the fencing used for width would be 2x, and the total length used for length would be y.
The sum of the lengths of the three sides would be the total available fencing:
2x + y = 24
We need to express 'y' in terms of 'x' to have a function of area.
Rearranging the equation:
y = 24 - 2x
The area of the rectangle can be found by multiplying the length and width:
Area = x * y
Substituting the value of 'y' from the equation above:
Area = x * (24 - 2x)
Now, we have the area as a function of 'x' only. To find the maximum area, we can differentiate the equation with respect to 'x' and find the value of 'x' at which the derivative equals zero.
d(Area)/dx = 24 - 4x
Setting the derivative equal to zero:
24 - 4x = 0
Solving for 'x':
4x = 24
x = 6
Now we can find the corresponding 'y' value using the equation y = 24 - 2x:
y = 24 - 2(6)
y = 24 - 12
y = 12
Therefore, the dimensions that result in the maximum area are a width of 6 and a length of 12. We can substitute these values into the area formula to find the maximum area:
Area = 6 * 12
Area = 72 square units
So, the maximum area that can be obtained using 24 meters of fencing with three sides and one wall is 72 square units.