A rectangular garden next to a building is to be fenced with 120 feet of fencing. The side against the building will not be fenced. What should the lengths of the other three sides be in order to assure the largest possible area?
2 w + L = 120 so L = (120-2w)
A = w L = w (120-2w) = 120 w - 2w^2
y = A/2 = 60 w - w^2
maximize A/2 same as max A
w^2 - 60 w = -y
w^2 - 60 w + 900 = -y + 900
(w-30)^2 = -(A/2) + 900
vertex at w = 30
the L = 120-60 = 60
so
30, 30 and 60
wow you rock thanks a bunch x
Well, this is a situation where we want to maximize the area of our rectangular garden, right? So, in order to do that, we should try to make the shape as close to a square as possible. After all, square is the new rectangle!
Since we know that one side is against the building and doesn't need fencing, let's call the lengths of the other three sides x, y, and z. Thus, we have:
x + y + z = 120
Now, we want to find the dimensions that maximize the area. The area of a rectangle is given by A = length × width. In this case, the length is x, and the width is the sum of the other two sides, y and z.
So, the area can be expressed as A = x(y + z). Since we want to maximize A, we should try to maximize x and (y + z).
In order to maximize x, we should make y and z as small as possible. This means that y = z = 0, so x + 0 + 0 = 120, which implies that x = 120.
Therefore, in order to assure the largest possible area, the side against the building should be 120 feet, and the other two sides should be 0 feet.
Congratulations, you've now created the world's first invisible garden! Just make sure you don't accidentally walk straight into it. Good luck! 🌱🦶
To find the dimensions that will result in the largest possible area, we need to consider the shape of the rectangle and the perimeter constraint.
Let's assume the length of the rectangle (parallel to the building) is represented by L, and the width (perpendicular to the building) is represented by W.
Given that the side against the building does not need to be fenced, we have two sides with length W and one side with length L that require fencing.
The perimeter of a rectangle is found by adding up the lengths of all sides:
Perimeter = 2W + L
From the information given, the perimeter is equal to 120 feet:
2W + L = 120
Since we want to find the largest possible area, we need to maximize L while still satisfying the perimeter constraint.
To proceed, let's solve the equation 2W + L = 120 for L in terms of W:
L = 120 - 2W
The area of the rectangle is given by the product of its length and width:
Area = L * W
Substituting the expression for L into the area equation:
Area = (120 - 2W) * W
To find the maximum area, we need to find the value of W that maximizes this equation. We can do this by taking the derivative of the equation with respect to W and setting it equal to zero.
Differentiating the area equation with respect to W:
d(Area)/dW = 120 - 4W
Setting the derivative equal to zero to find the maximum:
120 - 4W = 0
4W = 120
W = 30
Therefore, the width of the rectangle should be 30 feet.
Now, we can substitute this value of W back into the equation for L to find its value:
L = 120 - 2W
L = 120 - 2(30)
L = 120 - 60
L = 60
Therefore, the length of the rectangle should be 60 feet.
To summarize, in order to assure the largest possible area with a perimeter of 120 feet, the width of the rectangle should be 30 feet and the length should be 60 feet.
To find the dimensions that will result in the largest possible area, we need to use calculus. Let's call the width of the garden x and the length y. We know that two sides of the garden, each of length y, will be fenced with a total of 120 feet of fencing.
Given this information, we can set up an equation to represent the length of the fence:
2y + x = 120
Now, we need to express the area of the garden in terms of one variable so that we can maximize it. The area (A) of a rectangle is given by the product of its length and width:
A = xy
To simplify the equation, solve the first equation for x:
x = 120 - 2y
Substitute this expression for x in the area equation:
A = y(120 - 2y)
Now, we have an equation for the area in terms of a single variable y. To find the value of y that maximizes the area, we can take the derivative of A with respect to y, set it equal to zero, and solve for y:
dA/dy = 120 - 4y
Set dA/dy equal to zero and solve it for y:
120 - 4y = 0
4y = 120
y = 30
We have found that y = 30. Now, substitute this value back into the equation x = 120 - 2y to find x:
x = 120 - 2(30) = 120 - 60 = 60
So, for the largest possible area, the lengths of the other three sides should be 60 feet, 30 feet, and 30 feet, respectively.