a rectangular pen is to be constructed alongside a barn using 120 feet of fencing. the barn will be used for one side of the pen. What should the dimensions of the pen be to maximize its area.
***Can not use calculus. This is a 9th grade question
Well, isn't that a tricky one for a math problem! We need to find the dimensions of the pen that will maximize its area, given that one side is already provided by the barn.
Let's assume the width of the pen is x feet. Since we know the barn will be used for one side, we only need to find the length of the pen now.
To maximize the area, we want to make sure we use up all the fencing. So, let's set up an equation:
2x + L = 120
Here, 2x represents the two sides perpendicular to the barn, and L represents the length. We need to solve for L.
L = 120 - 2x
Now, we can calculate the area of the pen:
A = x * L
Since we know L in terms of x, we can substitute it into the equation:
A = x * (120 - 2x)
This simplifies to:
A = 120x - 2x^2
To find the maximum area, we can graph this equation and identify the vertex. But since we can't use calculus, let's think logically.
Since the coefficient of x^2 is negative (-2), we know the graph will be an upside-down parabola. So, the vertex will be at the maximum point.
The x-coordinate of the vertex in a quadratic function is given by -b/2a. In our case, a is -2 and b is 120:
x = -120 / (2 * -2)
x = 30
Now that we have x, we can find L:
L = 120 - 2x
L = 120 - 2 * 30
L = 60
So, the dimensions of the pen that will maximize its area are 30 feet by 60 feet. Voila!
To maximize the area of the rectangular pen, we need to determine the dimensions that will use up all 120 feet of fencing while still maximizing the area.
Let's assume the length of the pen extends away from the barn and the width of the pen extends parallel to the barn.
Let L be the length of the pen and W be the width of the pen.
Given that the barn is one of the sides of the pen, the total amount of fencing used will be the perimeter of the pen minus the length of the barn.
Perimeter = 2 * (Length + Width)
120 = 2 * (L + W)
Simplifying the equation, we have:
60 = L + W
Now, we need to express the area of the pen in terms of one variable, either L or W. The area of a rectangle is given by:
Area = Length * Width
Solving the equation L + W = 60 for L, we have L = 60 - W.
Substituting L = 60 - W into the area equation, we get:
Area = (60 - W) * W
Now, we need to find the Maximum Area.
To do so, we can create a table of values or sketch a graph to observe the trend.
Width (W) | Length (L = 60 - W) | Area (A = L * W)
---------------------------------------------------------
55 | 5 | 275
50 | 10 | 500
45 | 15 | 675
... | ... | ...
0 | 60 | 0
From the table above, we can see that as the width decreases, the length increases, resulting in a larger area. However, at a certain point, the width will become 0, resulting in an area of 0.
Therefore, the dimensions of the pen that maximize its area are:
Length = 60 - Width
Width = 0
In this case, the rectangular pen would be 60 feet long and 0 feet wide, which is essentially just a line perpendicular to the barn.
It's important to note that this solution demonstrates the maximum area achievable using the given constraints, assuming the pen can be any shape, including unrealistic ones. However, in practical scenarios, a pen with zero width would not be feasible for confinement purposes.
To maximize the area of the rectangular pen, we need to find the dimensions that will give us the largest possible area. We know that the pen will have three sides, and the fourth side will be formed by the barn.
Let's assume the length of the pen (parallel to the barn) is x feet. In this case, the remaining two sides will each be (120 - 2x)/2 = (60 - x) feet long.
The area of the rectangular pen is given by the formula: Area = Length x Width.
So, the area of the pen is A = x*(60 - x).
To find the dimensions that maximize the area, we can plot a graph of this equation. However, since calculus cannot be used, we can use a technique called "completing the square."
First, let's simplify the equation by expanding the expression:
A = 60x - x^2
Now, let's complete the square by adding and subtracting (60/2)^2 = 900:
A = -x^2 + 60x + 900 - 900
A = (-(x^2 - 60x + 900)) - 900
A = -(x - 30)^2 + 900 - 900
A = -(x - 30)^2
Since a negative sign is in front of the squared term, the area will be maximized when (x - 30)^2 is equal to zero. This happens when x = 30.
So, the length of the pen should be 30 feet for the area to be maximized. The width of the pen will be (60 - 30) = 30 feet as well.
Therefore, the dimensions for maximizing the area of the rectangular pen would be 30 feet by 30 feet.
If the sides have length a and b, then, assuming b is the side parallel to the barn,
a+a+b = 120, so b = 120-2a
Area = a*b = a*(120-2a) = 120a - 2a² = 2a(60-a)
Now, this is a parabola opening downward. The vertex (a maximum) is halfway between the roots, 0 and 60, at a=30
So, pen is 30x60 feet.