You have 7L feet of fence to make a rectangular vegetable garden alongside the wall of your house, where L is a positive constant. The wall of the house bounds one side of the vegetable garden. What is the largest possible area for the vegetable garden?
Let the side parallel to the house-wall by y
let the other two sides of the rectangle be x
so y + 2x = 2L
y = 2L - 2x
area = xy = x(2L - 2x) = -2x^2 + 2Lx
the maximum area occurs at the vertex of this downward parabola
the x of the vertex is -b/(2a) = -2L/-4 = L/2
so that max is -2(L/2)^2 + 2L(L/2) = .....
i
Who knew vegetables could be so demanding? Well, my green-thumbed friend, when it comes to maximizing the area of your rectangular vegetable garden, let's put on our thinking caps. Since you have 7L feet of fence, we know that the sum of all the sides of the rectangle should be equal to 7L.
Now, since one side is already taken care of by the wall of the house, we only need to figure out how to maximize the area with the remaining fence. Let's call the length of the rectangular garden L1 and the width L2.
The perimeter formula for a rectangle is 2L1 + 2L2, and we know that 2L1 + L2 = 7L. Solving for L2, we get L2 = 7L - 2L1.
Now, to maximize the area, we want to make L1 and L2 as close in value as possible while still satisfying the perimeter equation. In this case, that means making L1 and L2 the same.
So, with our newfound knowledge, we can set L1 = L2 and substitute it back into the equation: 2L1 + L1 = 7L. Simplifying, we get 3L1 = 7L, or L1 = 7L / 3.
Since the area of a rectangle is length times width, the area of the rectangular vegetable garden is L1 * L2 = L1^2. Plugging in L1 = 7L / 3, we get (7L / 3)^2 = 49L^2 / 9.
So, my veggie-loving friend, the largest possible area for your vegetable garden given 7L feet of fence is 49L^2 / 9. I hope your vegetables grow as big and fabulous as this answer! Happy gardening!
To find the largest possible area for the vegetable garden, we can use the formula for the area of a rectangle: A = length × width.
Let's assume the length of the garden is L (since we are given that L is a positive constant) and the width is W.
Since the wall of the house bounds one side of the garden, we have 3 sides to fence: two sides of length W and one side of length L.
Therefore, the total length of fence used is 2W + L.
Given that the total length of fence is 7L, we can set up the equation:
2W + L = 7L
Simplifying the equation, we get:
2W = 6L
W = 3L
Now we can substitute this value of W into the area equation:
A = L × W
A = L × 3L
A = 3L^2
As we want to find the largest possible area, we need to maximize A.
Taking the derivative of A with respect to L and setting it equal to zero, we can find the critical points.
dA/dL = 6L
Setting 6L equal to zero, we get:
6L = 0
L = 0
However, L cannot be zero as it is a positive constant.
Thus, there are no critical points and we can conclude that the maximum area occurs at the bounds of the given range (L being positive).
Therefore, the largest possible area for the vegetable garden is 3L^2, where L is a positive constant.
To find the largest possible area for the vegetable garden, we first need to determine the dimensions of the rectangular garden that will maximize the area.
Let's assume the length of the garden is L and the width is W. Since the wall of the house bounds one side of the garden, we can calculate the total length of all four sides of the garden:
2L + W = 7L (Equation 1)
Simplifying Equation 1, we get:
W = 5L (Equation 2)
We want to maximize the area of the garden, which is given by the formula:
Area = Length x Width = L x W
Substituting Equation 2 into the formula above, we have:
Area = L x (5L) = 5L^2
To maximize the area, we need to find the maximum value of 5L^2.
To do this, we take the derivative of the area function with respect to L, set it equal to zero, and solve for L:
d/dL (Area) = 10L = 0
Solving for L, we get:
L = 0
However, since L is a positive constant, the only valid solution for L is:
L = 0
Now, let's find the corresponding value of W:
W = 5L
Substituting L = 0, we have:
W = 5(0) = 0
Therefore, the dimensions of the garden are L = 0 and W = 0. However, this implies that the garden has no area, which is not possible.
Upon further analysis, we can see that as L approaches 0, the area of the garden becomes infinitesimally small. However, the moment we start increasing L, the area will increase as well.
Therefore, in order to maximize the area, we need to make L as large as possible while still satisfying the constraint given by the total length of the fence.
Since L is a positive constant, there is no maximum value for L. However, we can make L very close to the maximum possible value, which would result in a very large area for the vegetable garden.
In conclusion, the largest possible area for the vegetable garden is achieved when L is very large, approaching the maximum value allowed by the length of the fence.