Farmer Brown wants to fence in a rectangular plot in a large field, using a straight rock wall that is already there as the north boundary. The fencing for the east and west sides of the plot will cost $3 a yard, but she needs to use special fencing, which costs $5 a yard, on the south side of the plot. if the area of the plot is to be 600 square yards, find the dimensions of the plot that will minimize the cost of the fencing.
a) find the equation to maximized or minimized.
b) finding the solution.
c)showing that your solution is an absolute maximum or minimum.
length = 2 W + L
area = W*L = 600 so L = 600/W
C = 3(2W) + 5 L
C = 6 W + 5(600/W) = 6 W +3000/W
minimize C
dC/dW = 0 = 6 - 3000/W^2
W^2 = 3000/6 = 500
W = 10 sqrt 5 = 22.36
then L = 26.83
is that max or min?
d^2 C/d W^2 = 3000/W^4
second derivative is always positive so curving up always, this is minimum cost
a) To minimize the cost of the fencing, we need to minimize the total cost function. Let's consider the length of the plot as x and the width as y. The cost of fencing the east and west sides is 3y, while the cost of fencing the south side is 5x.
The area of the plot is given by xy = 600, so we can express y in terms of x: y = 600/x.
The total cost function, C(x), can be written as:
C(x) = 3y + 5x = 3(600/x) + 5x.
b) To find the solution, we need to find the value of x that minimizes C(x). We can do this by taking the derivative of C(x) with respect to x and setting it equal to zero:
C'(x) = -1800/x^2 + 5 = 0.
Solving this equation, we find:
1800/x^2 = 5.
1800 = 5x^2.
x^2 = 360.
x = √(360) or approximately 18.97.
Since x represents the length of the plot, it cannot be negative, so we take x = 18.97.
c) To show that this solution is an absolute minimum, we can use the second derivative test. Taking the second derivative of C(x):
C''(x) = 3600/x^3.
Since this is positive for all positive x, it means that C(x) is concave up. Therefore, the value we found for x (18.97) corresponds to the absolute minimum of the cost function C(x).
Therefore, the dimensions of the plot that will minimize the cost of the fencing are approximately 18.97 yards for the length and 600/18.97 yards for the width.
a) To find the dimensions that will minimize the cost of the fencing, we need to minimize the cost function. The total cost of the fencing is given by the equation:
Cost = 3(Length + Width) + 5(Width)
We need to express the dimensions of the plot in terms of a single variable to form the cost equation. Let's assume the width of the plot is x yards, and the length of the plot is y yards.
b) The area of the plot is given as 600 square yards, so we have:
Length * Width = 600
Substituting y = 600/x into the cost equation, we get:
Cost = 3(x + 600/x) + 5x
To find the minimum cost, we take the derivative of the cost function with respect to x and set it equal to zero:
d(Cost)/dx = 0
d/dx [3(x + 600/x) + 5x] = 0
Simplifying and solving for x, we get:
3 - 180000/x² + 5 = 0
-180000/x² + 8 = 0
180000/x² = 8
x² = 180000/8
x² = 22500
x = √22500
x ≈ 150
So, the width of the plot is approximately 150 yards.
To find the length, we substitute this value of x into the area equation:
Length = Area / Width
Length = 600 / 150
Length = 4
Therefore, the dimensions of the plot that will minimize the cost of the fencing are 150 yards by 4 yards.
c) To show that this solution is an absolute minimum, we can check the second derivative of the cost function. We take the second derivative and evaluate it at x = 150:
d²(Cost)/dx² = -360000/x³
d²(Cost)/dx² at x = 150 is approximately -360000/(150)³ < 0
Since the second derivative is negative, this confirms that the solution at x = 150 is a local minimum. Since there are no other critical points, it is also the absolute minimum. Hence, the dimensions of the plot minimize the cost of the fencing.
a) To minimize the cost of fencing, we need to find the dimensions of the plot that minimize the total cost of the fencing. Let's consider the length of the plot as x yards and the width as y yards.
The cost of the east and west sides (perimeter) of the fence is given as $3 per yard, which means the cost for both sides combined is $6 per yard. So, the cost for the east and west sides can be represented as 6x.
The cost of the special fencing on the south side is given as $5 per yard. Since the south side is the length of the plot, the cost for the south side can be represented as 5y.
The total cost of the fencing can be calculated by summing up the costs for the east and west sides and the south side:
Total cost = Cost for east and west sides + Cost for south side
Total cost = 6x + 5y
b) We are given that the area of the plot is 600 square yards. The formula for the area of a rectangle is A = length * width. So, we have the equation:
600 = xy
To find the dimensions of the plot that minimize the cost of the fencing, we can use the area equation to express one variable in terms of the other and substitute it into the total cost equation.
From the area equation, we can solve for y:
y = 600 / x
Substituting this value of y in the total cost equation, we get:
Total cost = 6x + 5(600 / x)
To find the minimum cost, we need to minimize the total cost function. We can do this by taking the derivative of the total cost function with respect to x, setting it equal to zero, and solving for x.
d(Total cost) / dx = 6 - 3000 / x^2 = 0
Now, solve for x:
6 - 3000 / x^2 = 0
3000 / x^2 = 6
x^2 = 3000 / 6
x^2 = 500
x = √500
x ≈ 22.36
Now, substitute this value of x back into the area equation to solve for y:
600 = 22.36 * y
y ≈ 26.85
Therefore, the dimensions that will minimize the cost of the fencing are approximately 22.36 yards for the length (x) and 26.85 yards for the width (y).
c) To show that this solution is an absolute minimum, we can take the second derivative of the total cost function with respect to x and check its sign.
d^2(Total cost) / dx^2 = 6000 / x^3
Since x is positive, the second derivative is positive. This indicates that the cost function has a minimum at x ≈ 22.36. Therefore, the solution we found is indeed an absolute minimum for the cost of the fencing.