A rectangular playground is to be fenced off and divided in two by another fence parallel to one side of the playground. 440 feet of fencing is used. Find the dimensions of the playground that maximize the total enclosed area. Remember to reduce any fractions and simplify your answers as much as possible.
-What is the shorter side of the playground? (220/3 sq ft)
-What is the longer side of the playground?
-What is the maximum area?
I know the shorter side is 220/3 sq ft, but I'm having difficulty getting the next two answers.
I will assume you made a diagram.
On my diagram, I have each of the long sides as y, and each of the 3 shorter sides as x
so we have 3x + 2y = 440
y = (440 - 3x)/2 = 220 - (3/2)x
area = xy
= x(220 - (3/2)x)
= 220x - (3/2)x^2
from the subject title, I assume you don't take Calculus, then this would be easy from here on, so ....
area = 220x - (3/2)x^2 is a downwards opening parabola, thus the area will have a maximum
we need the vertex.
Easiest way to get the vertex,
the x of the vertex is -b/(2a)
= -220/-3
= 220/3 , which you had but it should be ft, not square feet
so if x = 220/3
y = 220 - (3/2)(220/3)
= 220 - 110
= 110
short side is 220/3 ft, longer side is 110 ft
area = xy
= (220/3)(110)
= 24200/3 or 8066 2/3 square feet
or appr 8066.7 ft^2
No, not in calculus yet - I will take that next.
Thank you so much for your help!
To find the longer side of the playground and the maximum area, we can set up equations and use optimization techniques.
Let's assume that the shorter side of the playground is x feet. Since the fence is parallel to the shorter side, there are two sides of length x.
The longer side of the playground can be represented as y, and since the fence is parallel to this side, there are two sides of length y.
To find the total length of the fence used, we need to consider all four sides of the playground as well as the dividing fence. So we have:
2x + 2y + y = 440
Simplifying, we get:
2x + 3y = 440 Equation (1)
Now, let's express the area of the playground in terms of x and y. The area is given by A = x * y.
To find the maximum area, we need to maximize A. So, we need to express A in terms of a single variable. We can use Equation (1) to solve for x in terms of y and substitute it into the area equation.
From Equation (1), we can simplify it as:
2x = 440 - 3y
x = (440 - 3y) / 2 Equation (2)
Substituting Equation (2) into the area equation:
A = [(440 - 3y) / 2] * y
A = (440y - 3y^2) / 2
A = 220y - (3y^2 / 2) Equation (3)
Now we have the area equation in terms of y. To find the maximum area, we can take the derivative of Equation (3) with respect to y and set it to zero:
dA/dy = 220 - (3y/2) = 0
220 = (3y/2)
y = (220*2) / 3
y = 440/3 Equation (4)
Now we can substitute this value of y back into Equation (2) to find x:
x = (440 - 3*(440/3)) / 2
x = (440 - 440) / 2
x = 0 Equation (5)
From Equation (4), we have the longer side y = 440/3 and from Equation (5), we have the shorter side x = 0.
Therefore, the longer side of the playground is 440/3 feet, the shorter side is 0 feet, and the maximum area is A = 0.
Note: Since the value of x is 0, it means that the dividing fence should be on the longer side of the playground. Thus, the playground cannot be divided equally by the fence.
To find the longer side of the playground, we need to set up an equation using the given information.
Let's assume the shorter side of the rectangle is x feet.
Since the fence is parallel to one side of the playground, there will be two equal lengths of x that will be fenced off.
The perimeter of the rectangular playground is given as 440 feet. The equation for the perimeter is as follows:
2x + 2y = 440, where y is the longer side of the playground.
Rearranging the equation, we have:
2y = 440 - 2x
y = (440 - 2x)/2
y = 220 - x
Now, we can substitute the value of y into the area formula of a rectangle to calculate the total enclosed area.
Area = Length x Width
Area = x * y
Area = x * (220 - x)
To find the value of x that maximizes the area, we need to find the vertex of the quadratic equation x * (220 - x). Since the coefficient of x^2 is negative, the vertex will give us the maximum value of the area.
To find the x-coordinate of the vertex, we use the formula:
x = -b / 2a
In our case, a = -1 and b = 220.
x = -220 / (2 * -1)
x = 110
So, the shorter side of the playground is x = 110 feet.
Now, we can find the longer side using the equation for y:
y = 220 - x
y = 220 - 110
y = 110
Therefore, the longer side of the playground is 110 feet.
To find the maximum area, substitute the values of x and y into the area formula:
Area = x * y
Area = 110 * 110
Area = 12100 sq ft.
Hence, the maximum enclosed area is 12100 square feet.
To summarize:
- The shorter side of the playground is 110 feet.
- The longer side of the playground is also 110 feet.
- The maximum enclosed area is 12100 square feet.