a farmer wants to build a rectangular garden with a river as one side.what is the area of the biggest garden that he can fence with 220 meters of fencing materials?
110*55 = ?
To find the area of the biggest garden that can be fenced with 220 meters of fencing material, we need to determine the dimensions of the rectangle.
Let's assume that the width of the garden is 'w' meters and the length of the garden is 'l' meters. Since one side of the garden is the river, the length would be the same as the length of the fencing material.
Since the perimeter of a rectangle is the sum of all its sides, we can set up the following equation:
2w + l = 220
To maximize the area, we can express one variable in terms of the other. Let's solve the equation for l:
l = 220 - 2w
The area (A) of the garden can be calculated by multiplying the length (l) by the width (w):
A = w * l
A = w * (220 - 2w)
To find the maximum area, we can maximize this equation. Let's differentiate the equation with respect to 'w' and find the critical points:
dA/dw = 220 - 4w
Setting dA/dw to zero and solving for 'w':
220 - 4w = 0
4w = 220
w = 55
So, the width of the garden that maximizes the area is 55 meters.
Substitute this value into the equation for l:
l = 220 - 2w = 220 - 2 * 55 = 220 - 110 = 110
Therefore, the length of the garden is 110 meters.
Finally, we can calculate the area of the garden:
A = w * l
A = 55 * 110
A = 6050 square meters
Hence, the area of the biggest garden that can be fenced with 220 meters of fencing material is 6050 square meters.
To find the area of the biggest rectangular garden that can be fenced with 220 meters of fencing material, we need to understand the relationships between the perimeter, dimensions, and area of a rectangle.
Let's assume the width of the garden is "w" meters, and the length, or the side along the river, is "L" meters.
For a rectangle, the perimeter is given by the formula: perimeter = 2w + 2L.
In this case, one side of the garden is formed by the river. Therefore, the fencing required is the sum of three sides: perimeter = w + 2L.
Given that the total amount of fencing material is 220 meters, we can set up the equation:
w + 2L = 220.
Now, we need to maximize the area of the garden, which is given by the formula: area = w * L.
To solve this problem, we can use substitution. Rearranging the equation w + 2L = 220, we get w = 220 - 2L.
Substituting w in the area equation, we have: area = (220 - 2L) * L.
To find the maximum area, we can take the derivative of the area equation with respect to L, set it to zero, and solve for L.
Derivative of the area equation: d(area)/dL = 220 - 4L.
Setting the derivative equal to zero: 220 - 4L = 0.
Solving for L: L = 220/4 = 55.
Now that we have the value of L, we can substitute it back into the equation for w: w = 220 - 2L = 220 - 2(55) = 110.
Therefore, the width (w) is 110 meters, and the length (L) is 55 meters.
Finally, we can calculate the area of the rectangular garden: area = w * L = 110 * 55 = 6050 square meters.
Hence, the biggest garden that can be fenced with 220 meters of fencing material has an area of 6050 square meters.
L=LENGHT W=WIDTH
L+2w=220
L=220-2W
LW=AREA
A=W(L)
A=W(220-2W)
A=-2W^2- 2W^2
A=-2(W^2-110W+55^2-55^2)
A=-2(W-55)^2+6050
W=55
L=220-55-55
=110