A rectangular lot is to be bounded by a fence on three sides and by a wall on the fourth side. Two kinds of fencing will be used with heavy duty fencing selling for $4 a foot on the side opposite the wall. The two remaining sides will use standard fencing selling for $3 a foot. What are the dimensions of the rectangular plot of greatest area that can be fenced in at a cost of $6600?
THANK YOU!!!
cost = 6 b + 4 L = 6600
area = b L
so
4 L = 6600 - 6 b
L = 1650 - 1.5 b
area = y = 1650 b - 1.5 b^2
that is a parabola opening down, the maximum will be at the vertex
find the vertex
1.5 x^2 -1650 x = -y
x^2 - 1100 x = -2 y/3
x^2 - 1100 x + 302500 = -2 y/3 + 302500
(x-550)^2 = -2 y/3 + 302500
x = b = 550
so L = 1650 - 1.5 (550)
L = 825
To find the dimensions of the rectangular plot with the greatest area, we need to use calculus optimization techniques.
Let's assume that the length of the rectangular plot is x and the width is y.
We know that the cost of the heavy-duty fencing is $4 per foot, and the cost of the standard fencing is $3 per foot.
The total cost of the fencing can be expressed as follows:
Cost = 4x + 3(x + 2y)
We also know that the total cost is $6600. Hence, we can write:
6600 = 4x + 3(x + 2y)
Now, let's express the area of the rectangular plot in terms of x and y.
Area = xy
To find the values of x and y that maximize the area, we need to maximize the function Area = xy, subject to the constraint 6600 = 4x + 3(x + 2y).
We can rewrite the constraint equation as:
6600 = 4x + 3x + 6y
6600 = 7x + 6y
Now, let's solve the constraint equation for y:
6y = 6600 - 7x
y = (6600 - 7x)/6
Substituting this expression for y into the area equation, we have:
Area = x((6600 - 7x)/6)
To find the maximum area, we need to find the critical points of the area function. We can do this by differentiating the area function with respect to x and setting the derivative equal to zero:
d(Area)/dx = (6600 - 7x)/6 - 7x/6 = 0
Multiplying through by 6:
6600 - 7x - 7x = 0
6600 - 14x = 0
-14x = -6600
x = 6600/14
x = 471.43
Now we can substitute this value of x back into the constraint equation to solve for y:
6600 = 7(471.43) + 6y
6600 = 3300 + 6y
3300 = 6y
y = 550
Therefore, the dimensions of the rectangular plot that maximize the area are x = 471.43 feet and y = 550 feet.
To find the dimensions of the rectangular plot with the greatest area, we need to express the problem as a mathematical equation and then solve it. Let's start by defining the variables:
Let's call the length of the rectangular plot L, and the width W.
The cost of the heavy-duty fencing on the side opposite the wall is $4 per foot, and the cost of the standard fencing on the other two sides is $3 per foot.
We are given that the total cost of the fencing is $6600.
Now, let's create an equation based on the given information:
The cost of the heavy-duty fencing on the side opposite the wall is 4L.
The cost of the standard fencing on the other two sides is 3(L + W).
Adding these costs together should equal $6600:
4L + 3(L + W) = 6600.
Simplifying the equation, we get:
4L + 3L + 3W = 6600,
7L + 3W = 6600.
To find the maximum area, we can express area A in terms of length L and width W:
A = L * W.
Since we've already expressed W in terms of L in the previous equation, we can substitute it into the area equation:
A = L * (6600 - 7L) / 3.
To find the dimensions of the rectangular plot that maximize the area, we need to find the value of L that maximizes A. We can do this by finding the critical points of the area equation, which occur where the derivative is equal to zero.
To find the derivative of A with respect to L, we can use the quotient rule:
dA/dL = (6600 - 7L) / 3 - L * 7 / 3,
dA/dL = (6600 - 7L - 7L) / 3,
dA/dL = (6600 - 14L) / 3.
Now, set the derivative equal to zero and solve for L:
(6600 - 14L) / 3 = 0.
6600 - 14L = 0,
14L = 6600,
L = 6600 / 14,
L = 471.43 (approximately).
Now that we have the value of L, we can substitute it back into the equation for A to find the maximum area:
A = L * (6600 - 7L) / 3,
A = 471.43 * (6600 - 7 * 471.43) / 3,
A ≈ 471.43 * (6600 - 3300) / 3,
A ≈ 471.43 * 3300 / 3,
A ≈ 16006.67.
Therefore, the dimensions of the rectangular plot of greatest area that can be fenced in at a cost of $6600 are approximately 471.43 feet for the length (L) and (6600 - 7 * 471.43) / 3 ≈ 457.14 feet for the width (W).