A rectangular plot of land is to be fenced in using two kinds of fencing. Two opposite sides will use heavy-duty fencing selling for $5 a foot, while the remaining two sides will use standard fencing selling for $2 a foot. What are the dimensions of the rectangular plot of greatest area that can be fenced in at a cost of $7000?
if the dimensions are x and y, then we have the cost function
5*2x+2*2y = 7000
5x+2y = 3500
the area a = xy = x*(3500-5x)/2 = 1750x - 5/2 x^2
da/dx = 1750 - 5x
so maximum area is when the expensive sides are x = 350 ft
and the cheap sides have length y = 875 ft
Let's denote the length of one heavy-duty fencing side as `x`, and the length of one standard fencing side as `y`.
The cost of the heavy-duty fencing, 2 sides, is $5 * 2 * x = $10x.
The cost of the standard fencing, 2 sides, is $2 * 2 * y = $4y.
The total cost is $10x + $4y, which is given as $7000.
Therefore, we have the equation:
10x + 4y = 7000.
To find the dimensions of the rectangular plot with the greatest area, we need to maximize the area.
The area of a rectangle is given by the formula: A = length * width.
In this case, the area (A) is given by: A = x * y.
We can rewrite the equation for the cost using y in terms of x:
4y = 7000 - 10x
y = (7000 - 10x) / 4
Substituting this value of y in the area equation, we get:
A = x * ((7000 - 10x) / 4) = (7000x - 10x^2) / 4.
To maximize the area, we need to find the critical points by taking the derivative of A with respect to x and setting it equal to zero:
dA/dx = 0
(7000 - 20x) / 4 = 0
7000 - 20x = 0
20x = 7000
x = 350.
So, x = 350.
Now, substitute this value of x back into the cost equation to find y:
10x + 4y = 7000
10 * 350 + 4y = 7000
3500 + 4y = 7000
4y = 7000 - 3500
4y = 3500
y = 875.
So, the dimensions of the rectangular plot of greatest area that can be fenced in at a cost of $7000 are: length = 350 feet, and width = 875 feet.
To determine the dimensions of the rectangular plot of greatest area that can be fenced in at a cost of $7000, we need to set up an equation based on the given information.
Let's assume the length of the rectangular plot is "l" and the width is "w".
To calculate the cost, we need to consider that two opposite sides will use heavy-duty fencing (at $5 a foot) and the remaining two sides will use standard fencing (at $2 a foot).
The cost of the heavy-duty fencing would be $5 multiplied by the length (l) and $5 multiplied by the width (w). So, the cost of the heavy-duty fencing would be 2l + 2w multiplied by 5, which can be written as: 10l + 10w.
The cost of the standard fencing would be $2 multiplied by the length (l) and $2 multiplied by the width (w). So, the cost of the standard fencing would be 2l + 2w multiplied by 2, which can be written as: 4l + 4w.
Adding these two costs together gives us the total cost, which should equal $7000:
10l + 10w + 4l + 4w = 7000
Simplifying the equation, we get:
14l + 14w = 7000
Dividing both sides of the equation by 14, we have:
l + w = 500
To find the dimensions of the rectangular plot with the greatest area, we can use the fact that the area of a rectangle is given by length multiplied by width:
Area = l * w
We can substitute the value of "w" from the equation we derived earlier:
Area = l * (500 - l)
To find the maximum area, we need to differentiate the equation with respect to "l" and find where the derivative equals zero. This will give us the critical points where the area is maximized.
Differentiating with respect to "l":
d(Area)/d(l) = 500 - 2l
Setting d(Area)/d(l) equal to zero:
500 - 2l = 0
Solving for "l":
2l = 500
l = 250
Now that we have the value of "l," we can find the corresponding "w" by substituting it back into the equation:
w = 500 - l
w = 500 - 250
w = 250
Therefore, the dimensions of the rectangular plot of greatest area that can be fenced in at a cost of $7000 are length (l) = 250 feet and width (w) = 250 feet.