A farmer has 110 metres of fencing to fence off a rectangular area. Part of one side is a wall of length 15m. Find the dimensions of the field that give the maximum area. Answers: length and width = 31,25m
Thank you so much for a huge help.
Since there is additional 15 m which is part of one side of the wall, we can say that the total fencing material is
110 + 15 = 125 m
Recall that perimeter of rectangle is just
P = 2(L + W)
where L is the length and W is width.
Substituting,
125 = 2(L + W)
62.5 = L + W, or
L = 62.5 - W
Now, recall that area of rectangle is
A = L*W
Since we have an expression for length, we can substitute it here,
A = (62.5 - W)(W)
Since we want area to be maximized, we differentiate Area with respect to Width and equate to zero (since at maximum, the slope is zero):
A = 62.5*W - W^2
dA / dt = 62.5 - 2W
0 = 62.5 - 2W
2W = 62.5
W = 31.25 m (width)
Substituting this to get the length,
L = 62.5 - 31.25
L = 31.25 m (length)
Hope this helps~ :3
*lol, sorry it shouldn't be dA/dt, rather
dA/dW = 62.5 - 2W
but it won't affect the answer anyway. (it's just that you might wonder where the variable t came from) ^^;
To find the dimensions of the field that give the maximum area, we can start by setting up an equation using the given information.
Let L be the length of the field and W be the width.
We are told that the farmer has 110 meters of fencing, which can be used to enclose the perimeter of the rectangle, except for a wall of length 15m. This means the total fencing needed is:
2L + W + W - 15
Simplifying this, we have:
2L + 2W - 15 = 110
Next, we can simplify the equation by isolating one variable. Let's solve for L:
2L = 110 - 2W + 15
2L = 125 - 2W
L = 62.5 - W
Now, we can find an equation for the area of the field. The area of a rectangle is given by multiplying its length and width, so:
Area = L * W
Substituting the value of L we found earlier, we have:
Area = (62.5 - W) * W
To find the dimensions that give the maximum area, we can differentiate the area function with respect to W and set it equal to zero.
d(Area)/dW = 0
Using the product rule, we differentiate the area expression:
d(Area)/dW = (62.5 - W) * 1 + (-1) * W
Simplifying this, we get:
0 = 62.5 - 2W
2W = 62.5
W = 31.25
Now, we can substitute this value back into the equation for L:
L = 62.5 - W
L = 62.5 - 31.25
L = 31.25
Therefore, the dimensions of the field that give the maximum area are length = 31.25m and width = 31.25m.
To find the dimensions of the field that give the maximum area, we need to set up an equation using the information provided.
Let's assume the length of the field is represented by 'L' and the width is represented by 'W'. From the given information, we know that one side of the field is a wall with a length of 15m.
Since we have a rectangular field with a wall on one side, the perimeter equation can be written as:
2L + W + 15 = 110 (1)
This equation represents the total length of fencing used.
We need to rearrange this equation to express one variable (L or W) in terms of the other, so we can substitute it into the area equation.
Subtracting 15 from both sides of equation (1), we get:
2L + W = 95 (2)
Next, let's express one variable in terms of the other. We'll solve equation (2) for W:
W = 95 - 2L (3)
The area of a rectangle is given by the equation:
A = L * W
Substituting equation (3) into the area equation:
A = L * (95 - 2L)
Expanding the expression:
A = 95L - 2L^2
Now, to find the maximum area, we need to find the value of L that maximizes the area. We can do this by finding the critical points of the area equation.
To find the critical point, we take the derivative of the area equation with respect to L and set it equal to zero. This will give us the value of L at the maximum area.
dA/dL = 95 - 4L
Setting dA/dL = 0:
95 - 4L = 0
Solving for L:
4L = 95
L = 95/4
L = 23.75
Now that we have found the value of L that maximizes the area, substitute it back into equation (3) to find the corresponding value of W:
W = 95 - 2L
W = 95 - 2(23.75)
W = 95 - 47.5
W = 47.5
Therefore, the dimensions of the field that give the maximum area are:
Length (L) = 23.75m
Width (W) = 47.5m
NOTE: The answers given, length = 31m and width = 25m, do not satisfy the given conditions or the maximization of the area. The correct dimensions are 23.75m and 47.5m.