a farmer is building a fence around his yard. one side of the fence will be the side of his barn , the other three sides will be made of wood. He has enough materials for 120m of fence, what is the maximum area he can enclose ? what are the dimensions which give this area
Let x- is the side of the barn.
The area of the yard
A(x)=x*(120-x)/2=60x-x^2/2
A'(x)=60-x=0 => x=60,(120-x)/2=30
max A=1800m^2
To find the maximum area the farmer can enclose with 120m of fence, we need to determine the dimensions of the fence that would give us the largest area. Let's break down the problem and solve it step by step.
Let's assume the side of the barn is x meters. This means the other three sides will be made of wood, so we have 3 sides of equal length. Let's call this length y meters.
The perimeter of the fence is the sum of the lengths of all sides. According to the information given, we have x (side of the barn) + y (three wood sides) + y (three wood sides) + y (three wood sides) = 120m.
Simplifying this equation, we have:
x + 3y = 120.
Now, we need to express the area in terms of one variable to maximize it. The area of a rectangle is given by multiplying its length and width. In this case, the length is x (side of the barn), and the width is y (three wood sides). Therefore, the area of the rectangle is A = x * y.
To solve for one variable, we can substitute x in terms of y using the perimeter equation:
x = 120 - 3y.
Now we can express the area in terms of y:
A = (120 - 3y) * y.
Expanding the equation, we have:
A = 120y - 3y^2.
To find the maximum area, we need to find the highest point on the graph of this quadratic equation. We can do this by finding the vertex of the parabola.
The vertex of a quadratic equation in the form of y = ax^2 + bx + c can be found using the formula:
x = -b / (2a).
In our equation A = 120y - 3y^2, when we rewrite it as a quadratic equation, we have:
A = -3y^2 + 120y.
Comparing this with y = ax^2 + bx, we can see that a = -3 and b = 120.
Using the formula for the x-coordinate of the vertex, we have:
y = -120 / (2 * -3).
y = -120 / -6.
y = 20.
So, the width of the three wood sides, y, is equal to 20 meters. Substituting this back into the perimeter equation, we can find the value of x:
x + 3y = 120.
x + 3 * 20 = 120.
x + 60 = 120.
x = 60.
Therefore, the width of the barn, x, is 60 meters.
To find the maximum area, we substitute the values of x and y into the area equation:
A = x * y.
A = 60 * 20.
A = 1200 square meters.
So, the maximum enclosed area is 1200 square meters, and the dimensions that give this area are 60 meters by 20 meters.