A farmer wants to fence a small rectangular yard next to a barn. Fence for side parallel to the barn will cost 75 per foot and the fence for the other two sides will cost 30 per foot. The farmer has a total of 1750 dollars to spend on the project. Find the dimensions for the yard that will have the largest possible area.
The side parallel to the barn should be feet long and the other two sides should be feet long each.
A = x y
cost = 75 x + 60 y = 1750
60 A = x (1750 - 75 x)
maximize 60 A = z
z = 1750 x -75 x^2
75 x^2 -1750 x = -z
complete square to find vertex o parabola
x^2 - 23.3 x = -z/75
x^2 - 23.3 x + 11.7^2 = - z/75 + 136
(x -11.7)^2 = -(1/75)(z-10,200)
x = length = 11.7
A = z/60 = 170
so x y = 11.7 y = 170
y = width = 14.5
Let's assume the side parallel to the barn has a length of x feet and the other two sides have lengths y feet each.
To find the dimensions that will maximize the area, we need to maximize the area function A(x, y).
The area of a rectangle is given by the formula A = length x width.
For this specific rectangle, the length is x feet and the width is y feet.
Therefore, A(x, y) = x * y.
Now, we need to consider the cost of the fence. The cost of the fence for the side parallel to the barn is $75 per foot, while the cost of the fence for the other two sides is $30 per foot.
The total cost of the fence is $1750.
The cost of the fence for the side parallel to the barn is 75x dollars, and the cost of the fence for the other two sides is 30y dollars each, so the total cost equation is:
75x + 30y + 30y = 1750
Simplifying this equation, we have:
75x + 60y = 1750
Dividing both sides by 5:
15x + 12y = 350
Now, we have two equations:
- A(x, y) = x * y
- 15x + 12y = 350
To find the dimensions that will maximize the area, we need to find the critical points of the function A(x, y) subject to the constraint.
To do this, we can solve the second equation for one variable and substitute it into the area equation.
Simplifying the second equation, we have:
15x = 350 - 12y
x = (350 - 12y)/15
Substituting this value of x into the area equation:
A(y) = [(350 - 12y)/15] * y
Now, we have the area function in terms of a single variable y.
To find the maximum value of A(y), we can take the derivative of the function with respect to y and set it equal to zero.
dA/dy = [(350 - 12y)/15] - (12/15)y
Setting this equal to zero, we have:
[(350 - 12y)/15] - (12/15)y = 0
Multiplying both sides by 15:
350 - 12y - 12y = 0
Simplifying:
24y = 350
y = 350/24
y = 14.58 (rounded to two decimal places)
Now, we can use this value of y to find the corresponding value of x:
x = (350 - 12y)/15
x = (350 - 12*14.58)/15
x = 350/15
x = 23.33 (rounded to two decimal places)
Therefore, the side parallel to the barn should be approximately 23.33 feet long, and the other two sides should be approximately 14.58 feet long each.
To find the dimensions for the yard that will have the largest possible area, we need to maximize the area function. Let's assume the side parallel to the barn has a length of x feet, and the other two sides have a length of y feet each.
The cost of the fence parallel to the barn is $75 per foot, so its cost will be 75x dollars.
The cost of the other two sides is $30 per foot, so their combined cost will be 2(30y) = 60y dollars.
Since the total budget for the project is $1750, we can set up the following equation:
75x + 60y = 1750
Now, we need to express the area in terms of x and y. The area of a rectangle is given by the length times the width, which in this case is x * y.
To maximize the area, we need to maximize the product xy.
Now, let's solve the equation:
75x + 60y = 1750
To make things easier, let's divide the equation by 5:
15x + 12y = 350
Now, let's solve for y in terms of x:
12y = 350 - 15x
y = (350 - 15x) / 12
Substitute this expression for y in the area equation:
A = xy = x[(350 - 15x) / 12] = (350x - 15x^2) / 12
To find the maximum area, we need to find the critical points of the area function. We can do this by taking the derivative and setting it equal to zero:
dA/dx = (350 - 30x) / 12 = 0
Simplifying the equation:
350 - 30x = 0
30x = 350
x = 350/30
x = 35/3
Now, we can substitute this value of x back into the equation for y:
y = (350 - 15(35/3)) / 12
y = (350 - 525/3) / 12
y = (350 - 175) / 12
y = 175/12
So, the dimensions for the yard that will have the largest possible area are:
The side parallel to the barn should be (35/3) feet long
The other two sides should be (175/12) feet long each.