farmer has 2400ft of fencing and wants to fence off a rectangular field that borders a straight river he does not need a fence along the river what are the dimensions of the feild of largest area that he can fence
Experiment with the problem drawing several diagrams illustrating the situation calculate the area of each configuration and use your results to estimate the dimensions of target possible field enter your answer as a comma separated list
Find a function that models the area of the feild in terms of one of its sides
A(x)
To find the dimensions of the field with the largest area that can be fenced, let's assume the length of the field is x feet.
Since the field is rectangular, we can divide the available fencing into 3 sides: two sides of length x feet each and one side along the river. The total length of these three sides should add up to 2400 feet, which gives us:
2x + river side = 2400
river side = 2400 - 2x
To find the area of the field, we use the formula: A = length * width. Since the width is the side opposite the river, it will be the remaining side of length x feet:
A = x * (2400 - 2x)
A = 2400x - 2x^2
To find the maximum area, we need to find the critical point of this quadratic equation by taking its derivative and setting it to zero:
A'(x) = 2400 - 4x = 0
4x = 2400
x = 600
So, the length of the field should be 600 feet. The width would be the remaining side, which is (2400 - 2x) = (2400 - 2*600) = 1200 feet.
Therefore, the dimensions of the field with the largest area that can be fenced are 600 feet by 1200 feet.
The function that models the area of the field in terms of one of its sides x is:
A(x) = 2400x - 2x^2
To find the dimensions of the field with the largest area, we can follow these steps:
1. Let's assume the width of the field is 'x' feet. Since there is no need to fence along the river, we only need three sides of the rectangle to be fenced.
2. The length of the field will be (2400 - 2x) feet, as we subtract the width from the total length of fencing.
3. The area of a rectangle is given by the formula A = length * width. Therefore, the area of the field can be calculated as A(x) = x * (2400 - 2x).
Now, we can simplify the expression A(x):
A(x) = x * (2400 - 2x)
A(x) = 2400x - 2x^2
To find the maximum area, we need to find the value of 'x' that maximizes the function A(x).
To do this, we can take the derivative of A(x) with respect to 'x' and set it equal to zero:
A'(x) = 2400 - 4x
2400 - 4x = 0
4x = 2400
x = 600
The critical point is x = 600.
To confirm that this is indeed the maximum, we can take the second derivative of A(x) and evaluate it at x = 600:
A''(x) = -4
Since the second derivative is negative, this confirms that x = 600 is the maximum.
Therefore, the dimensions of the field with the largest area are width = 600 feet and length = 2400 - 2(600) = 1200 feet.
So, the answer is: 600, 1200.
To solve this problem, we need to find the dimensions of the rectangular field that will maximize its area while using a total of 2400ft of fencing.
Let's start by drawing a diagram of the problem. We'll represent the length of the field with 'l', and the width with 'w'. The river will be at one end of the rectangle, so we only need to fence the other three sides.
l
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w| |
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Since we know that the total length of the fencing is 2400ft, we can express this as an equation: 2l + w = 2400. This accounts for two lengths and one width, as there is no need to fence along the river.
Next, let's find a function that models the area of the field in terms of one of its sides. The area of a rectangle is given by the formula A = l * w.
We can rearrange the equation 2l + w = 2400 to solve for w: w = 2400 - 2l. Now we can substitute this value of w into the area equation: A = l * (2400 - 2l).
Simplifying further, we get A = 2400l - 2l^2. This is the function that models the area of the field in terms of l, the length of the field.
To find the dimensions of the field that maximize its area, we need to find the value of l that maximizes the function A(l). We can do this by analyzing the critical points of the function.
To find the critical points, we take the derivative of A(l) with respect to l, set it equal to zero, and solve for l. Let's differentiate A(l):
A'(l) = 2400 - 4l
Setting A'(l) = 0 and solving for l:
2400 - 4l = 0
4l = 2400
l = 600
Now we have found a critical point at l = 600. We need to ensure this is a maximum and not a minimum. We can do this by taking the second derivative of A(l) and evaluating it at l = 600.
A''(l) = -4
Since A''(l) is negative, this confirms that the critical point at l = 600 is a maximum point. Therefore, when the length of the field is 600ft, it will maximize the area.
Now, we can substitute this value of l back into the equation 2l + w = 2400 to find w:
2(600) + w = 2400
1200 + w = 2400
w = 1200
Thus, the dimensions of the field of largest area that can be fenced with 2400ft of fencing are 600ft by 1200ft.
So, the answer is: 600, 1200.