a gardner having 120m of fencing whishes to enclose a rectangular plot of land and also a erect a fence across the land parellel to two of the sides. find the maximum area he can enclose
3 W + 2 L = 120 ... L = 60 - 3/2 W
a = L W ... substituting ... a = 60 W - 3/2 W^2
the max is on the axis of symmetry
... Wmax = -60 / (2 * -3/2) = 20
substitute back to find Lmax , then calculate max area
Ah, the gardener's dilemma! Well, to maximize the area, the shape should be a square. Let me explain why in a clownish manner.
Imagine the gardener's plot of land is like a delicious pie, and the fencing is the crust. Now, if the gardener creates a rectangle, some of the delicious filling will be left out! But if the gardener creates a square, it will perfectly enclose the filling with minimum crust.
So, if the gardener has 120 meters of fencing, they would use 30 meters for each side of the square (since all sides are equal in a square). This means the maximum area they can enclose is 30 meters multiplied by 30 meters, which is a whopping 900 square meters! That's a whole lot of plant-growing space!
Remember, in gardening and jokes, always aim for the "squarest" solution for maximum enjoyment and laughter!
To find the maximum area that the gardener can enclose, we need to determine the dimensions of the rectangular plot of land.
Let's assume the length of the rectangle is 'L' and the width is 'W'.
According to the problem, the gardener has 120 meters of fencing. This means that the total length of the fencing used will be the perimeter of the rectangle.
The perimeter of a rectangle is given by the formula:
Perimeter = 2L + 2W
In this case, the perimeter is 120 meters:
2L + 2W = 120
We can simplify this equation by dividing both sides by 2:
L + W = 60
Now, we need to express the area of the rectangle, 'A', in terms of 'L' and 'W'.
The area of a rectangle is given by the formula:
Area = L * W
We want to maximize the area, so we need to maximize the product of 'L' and 'W'. Since we already have an equation for 'L' in terms of 'W', we can substitute it into the area formula:
Area = L * (60 - L)
To maximize the area, we need to find the maximum value of this equation.
We can do this by finding the maximum point of the quadratic equation. Taking the derivative of the equation with respect to 'L' and setting it equal to zero will give us the value of 'L' that maximizes the area.
d/dL (Area) = 60 - 2L = 0
Now, solve for 'L':
60 - 2L = 0
2L = 60
L = 30
So, the length that maximizes the area is 30 meters.
Substituting this value of 'L' back into the equation for the perimeter, we can find the value of 'W':
30 + W = 60
W = 60 - 30
W = 30
So, the width that maximizes the area is also 30 meters.
Finally, we can calculate the maximum area:
Area = L * W
Area = 30 * 30
Area = 900 square meters
Therefore, the gardener can enclose a maximum area of 900 square meters.
To find the maximum area that can be enclosed, we need to determine the dimensions of the rectangular plot.
Let's assume the length of the rectangular plot is x meters. In this case, the width of the plot will be 120 - 2x meters (as we need to subtract twice the length of the fence from the total available fencing).
The area of a rectangle can be calculated by multiplying the length by the width. So the area A(x) in terms of x is:
A(x) = x(120 - 2x)
To find the maximum area, we need to find the value of x that maximizes this function.
To do this, we can take the derivative of A(x) with respect to x and set it to zero:
A'(x) = 120 - 4x
120 - 4x = 0
Solving for x:
4x = 120
x = 30
So the length of the rectangular plot that would maximize the area is 30 meters. Substituting this back into the equation for the width:
Width = 120 - 2x = 120 - 2(30) = 60 meters
Therefore, the maximum area that can be enclosed is:
A(max) = Length × Width = 30 × 60 = 1800 square meters.