A farmer decides to enclose a rectangular garden, using the side of a barn as one side of the rectangle. What is the maximum area that the farmer can enclose with 80 feet of fence? What should the dimensions of the garden be to give the area?
He will need one length and two widths
let the length by y ft
let the width be x ft
2x + y = 80
y = 80-2x
area = xy
= x(80-2x)
= -2x^2 + 80x
The x value to produce the maximum area is the x of the vertex of this parabola.
If you know Calculus
d(area)/dx = -4x + 80= 0 for a max of area
x = 20
then y = 80-2x = 40
the field should be 20 by 40 , with 40 being the one side.
non-Calculus:
the x of the vertex of y = ax^2 + bx +c is -b/(2a)
which in this case is -80/(2(-2)) = 20
continue as above.
or you can complete the square:
area = -2(x^2 - 40x + 400 - 400 )
= -2(x-20)^2 + 800
max of area is 800 when x = 20 and y = 40
The farmer wants to use the 500 m to an enclosure divided into two equal areas. What is the total maximum area that can is achieved with the 500 m fence. use differential calculus to the solution.
To find the maximum area that the farmer can enclose with 80 feet of fence, we need to consider the formula for the perimeter of a rectangle.
Let's assume the length of the garden is x and the width is y. Since the side of the barn is one side of the rectangle, the fence will only be needed for three sides.
The formula for the perimeter of a rectangle is: P = 2x + 2y
Given that the perimeter is 80 feet, we can set up the equation: 2x + 2y = 80
Simplifying the equation, we get: x + y = 40
To find the maximum area, we need to maximize the length and width of the rectangle.
Since one side is given by the barn, the length of the rectangle cannot be more than 40 feet. Thus, we can assume x = 40.
Substituting x = 40 in the equation x + y = 40, we get: 40 + y = 40
y = 40 - 40
y = 0
Therefore, the dimensions of the garden should be 40 feet by 0 feet. However, having a width of 0 feet would make it just a line, not a rectangle.
So, we need to consider the scenario where the length of the rectangle is less than 40 feet. Assuming the length is less than 40 feet, let's say x = 20.
Now, substituting x = 20 in the equation x + y = 40, we get: 20 + y = 40
y = 40 - 20
y = 20
We now have dimensions of the garden: 20 feet by 20 feet.
To find the area, we multiply the length and width:
Area = length × width
Area = 20 ft × 20 ft
Area = 400 square feet.
Therefore, the maximum area that the farmer can enclose with 80 feet of fence is 400 square feet, and the dimensions of the garden should be 20 feet by 20 feet.
To find the maximum area that the farmer can enclose with 80 feet of fence, we need to consider the different dimensions of the garden.
Let's assume the side of the barn is one of the shorter sides of the rectangle, and the other two sides are of equal length (let's call them x). The total length of the three sides would be x + x + 80 = 2x + 80.
To find the dimensions that would give the maximum area, we need to find the value of x that maximizes the area of the rectangle.
The area of a rectangle is given by length multiplied by width. In this case, the length is the side of the barn (call it y), and the width is x. So, the area of the rectangle is A = y * x.
Now, let's express the length of the other two sides in terms of x and the side of the barn. The length of those two sides would be y + y = 2y.
Since we have the total length of the three sides equal to 2x + 80, we can express y in terms of x: 2y = 2x + 80 - y. Simplifying further, we get y = (2x + 80) / 2 = x + 40.
Substituting this value of y back into the equation for the area, we get A = (x + 40) * x = x^2 + 40x.
Now, we have an equation for the area in terms of x. To find the maximum area, we can take the derivative of this equation with respect to x, set it equal to zero, and solve for x.
dA/dx = 2x + 40
Setting this equal to zero, we have:
2x + 40 = 0
2x = -40
x = -20
Since we are looking for positive dimensions, we can discard the negative value of x.
Therefore, the dimensions of the garden that would give the maximum area are x = -20 feet (discarded) and y = x + 40 = -20 + 40 = 20 feet.
So, the maximum area that the farmer can enclose with 80 feet of fence is given by A = x * y = 20 * -20 = 400 square feet.
The dimensions of the garden to achieve this maximum area would be 20 feet by 20 feet.