A rectangular play yard is to be constructed along the side of a house by erecting a fence on three sides, using house wall as the fourth wall. Find the demensions that produce the play yard of maximum area if 20 meters of fence is available for the project.
Make a sketch,
let the width of the field be x,
let the length of the field be y
then y + 2x = 20 , (we need only one length)
y = 20-2x
area = xy
= x(20-2x)
= -2x^2 + 20x
I don't know at what level of math-study you are.
If you know calculus, find
d(area)/dx = -4x + 20 = 0
x = 5 , then y = 10 for a max area of 50
If you don't know calculus, complete the square of the quadratic function
A = -2x^2 + 20x
= -2(x^2 - 10x + 25 - 25 )
= -2((x-5)^2 - 25)
= -2(x-5)^2 + 50
vertex is (5,50)
so the max area is 50 when x = 5 , as above
To find the dimensions that produce the play yard of maximum area, we need to determine the length and width of the rectangular play yard.
Let's assume the length of the yard is L and the width is W.
Given that we have 20 meters of fence available, we can determine the perimeter of the rectangular play yard.
The perimeter is equal to the sum of all four sides of the rectangle:
Perimeter = L + W + L = 2L + W = 20 meters
Now, let's express the width W in terms of L:
W = 20 - 2L
Next, we need to find the area of the play yard. The area of a rectangle is given by:
Area = Length * Width
Substituting the expression for W:
Area = L * (20 - 2L)
To find the maximum area, we can take the derivative of the area function with respect to L and set it equal to zero.
d(Area)/dL = 20 - 4L = 0
Solving for L:
4L = 20
L = 20/4
L = 5 meters
Substituting the value of L back into the expression for W:
W = 20 - 2L
W = 20 - 2(5)
W = 20 - 10
W = 10 meters
Therefore, the dimensions of the play yard that produce the maximum area with 20 meters of fence are 5 meters by 10 meters.
To find the dimensions that produce the play yard of maximum area, we can use the concept of optimization.
Let's assume the width of the play yard is x meters. Since the play yard is rectangular, the length of the play yard will be 20 - 2x meters, as we need to subtract the two sides that are adjacent to the house wall.
The area of a rectangle is given by the formula A = length × width. So, in this case, the area of the play yard is A = (20 - 2x) × x.
To maximize the area, we can differentiate the area function with respect to x and find the value of x that makes the derivative equal to zero.
Differentiating A = (20 - 2x) × x with respect to x gives:
dA/dx = (20 - 4x)
Setting dA/dx = 0, we have:
20 - 4x = 0
Solving for x, we find:
4x = 20
x = 5
Therefore, the width of the play yard is 5 meters.
Substituting this value back into the equation for the length, we can find the length:
length = 20 - 2x = 20 - 2(5) = 10 meters
So, the dimensions that produce the play yard of maximum area are a width of 5 meters and a length of 10 meters.