Sarah has 20 m of garden edging. What are the dimensions of the rectangular garden with the greatest area can she enclose with the edging? What is the maximum area?
the greatest area is when the rectangle is a square. That would mean each side is 20/4 = 5m
The maximum area is 5^2 = 25 m^2
Multiply any two other two numbers that add up to 10 and you will get less than 25.
Well, Sarah is in a bit of a "circum-stance" here. Since she has a limited amount of garden edging (20 m), we'll have to find the optimal dimensions to maximize the garden area.
Let's assume the rectangular garden has length L and width W. We know that the perimeter of a rectangle is given by the formula P = 2L + 2W. In this case, P = 20 m.
Since Sarah's total available edging is 20 m, that means 2L + 2W = 20. We can simplify this to L + W = 10, or W = 10 - L.
Now, to maximize the garden area, we need to find the largest possible product of L and W. Let's multiply them together: A = L * W = L * (10 - L) = 10L - L^2.
To find the maximum area, we need to find the vertex of this quadratic equation, given by the formula -b / (2a). In this case, a = -1, b = 10.
Calculating, we get L = -10 / (2 * -1) = 5.
So the dimensions of the rectangular garden with the greatest area that Sarah can enclose with the 20 m of edging are 5 m by 5 m.
And the maximum area is obtained by substituting L = 5 into our equation: A = 5 * (10 - 5) = 25 square meters.
So Sarah can enclose a garden with dimensions 5 m by 5 m, giving her a maximum garden area of 25 square meters.
To find the dimensions of the rectangular garden with the greatest area that can be enclosed with the given 20 m of garden edging, we will use the concept of optimization.
Let's assume the length of the garden is L and the width is W.
We are given that the total garden edging is 20 m, so the perimeter of the rectangular garden can be expressed as:
Perimeter = 2L + 2W
Since the total garden edging is 20 m, we have:
2L + 2W = 20
Now, we need to express the area of the rectangular garden in terms of L and W. The area of a rectangle is given by the formula:
Area = length × width
Therefore, for the rectangular garden, the area can be expressed as:
Area = L × W
To maximize the area, we need to maximize the product of L and W.
To solve this problem, let's use the equation from the perimeter:
2L + 2W = 20
Solving for L, we can express L in terms of W:
2L = 20 - 2W
L = (20 - 2W) / 2
L = 10 - W
Substituting this value of L into the equation for the area, we get:
Area = (10 - W) × W
Area = 10W - W²
To find the maximum area, we can take the derivative of the area function with respect to W and set it equal to zero:
d(Area) / dW = 0
Differentiating the area equation, we get:
d(Area) / dW = 10 - 2W
Setting this equal to zero:
10 - 2W = 0
Solving for W:
2W = 10
W = 10 / 2
W = 5
Now that we have the value of W, we can substitute it back into the equation for L:
L = 10 - W
L = 10 - 5
L = 5
So, the width of the rectangular garden with the greatest area is 5 m, and the length is also 5 m.
Therefore, the dimensions of the rectangular garden with the greatest area that can be enclosed with the given garden edging is 5 m by 5 m.
To find the maximum area, we can substitute the values of L and W into the area equation:
Area = L × W
Area = 5 × 5
Area = 25 square meters
Thus, the maximum area that can be enclosed with the given garden edging is 25 square meters.
To find the dimensions of the rectangular garden with the greatest area that Sarah can enclose with 20m of garden edging, we can use calculus. Let's assume that the width of the rectangular garden is x meters.
- Step 1: Express the perimeter of the rectangular garden in terms of x.
- The perimeter is given by P = 2(length + width) = 2(x + length).
- Since we know that the perimeter is 20m, we can write this as 20 = 2(x + length), or x + length = 10.
- Step 2: Express the area of the rectangular garden in terms of x.
- The area is given by A = length * width = length * x.
- We can rewrite this using the relationship x + length = 10 as A = x(10 - x) = 10x - x^2.
- Step 3: Find the derivative of the area function to find critical points.
- We differentiate the area function A = 10x - x^2 with respect to x: dA/dx = 10 - 2x.
- Step 4: Set the derivative equal to zero and solve for x to find the critical points.
- Setting 10 - 2x = 0, we find that x = 5.
- Step 5: Determine the maximum area.
- To determine whether the critical point is a maximum or minimum, we can take the second derivative of the area function.
- The second derivative is d^2A/dx^2 = -2.
- Since the second derivative is negative, the critical point x = 5 corresponds to a maximum.
Therefore, the width of the rectangular garden is x = 5 meters, and the length is 10 - x = 10 - 5 = 5 meters. Hence, the dimensions of the rectangular garden with the maximum area that can be enclosed with 20m of garden edging are 5m x 5m.
To calculate the maximum area, substitute x = 5 into the area function: A = 10(5) - (5)^2 = 50 - 25 = 25 square meters.
So, the maximum area that Sarah can enclose with 20m of garden edging is 25 square meters.