I want to make a rectangular garden by fencing in a area on the back of my house. Three sides of the rectangle will be fencing; the fourth side will be the side of the house. I have 60 feet of fence. What dimensions should I make the rectangle so that I have the largest possible area for the garden?
W = The width of fence
L = Length of house = third side of fence
2 W + L = 60 ft
L = 60 - 2 W
A = W * L
A = W * ( 60 - 2 W )
A = 60 W - 2 W ^ 2
Now you must use calculus :
If a first derivative of a function at some point is equal to zero function has maxsimum or minimum.
If the second derivative is negative then the function has maximum.
If the second derivative f'' is positive then the function has minimum.
First derivative:
d A / d W = 60 - 2 * 2 W = 60 - 4 W
d A / d W = 60 - 4 W = 0
60 = 4 W Divide both sides by 4
15 = W
W = 15 ft
L = 60 - 2 W
L = 60 - 2 * 15
L = 60 - 30
L = 30 ft
Second derivative = - 4
the function has maximum.
Maximum area :
Amax = W * L = 15 * 30 = 450 ft ^ 2
thank you
To find the dimensions that will give you the largest possible area for the garden, you can use the concept of optimization. In this case, you need to maximize the area of the garden while using a fixed length of 60 feet for the fencing.
Let's assume the width of the rectangle is x feet. Since you have three sides of fencing, the length of the other two sides would be (60 - x - x) = (60 - 2x) feet.
The area of a rectangle is given by the formula: Area = Length × Width.
So, in this case, the area of the garden would be: Area = (60 - 2x) × x.
To find the dimensions that maximize the area, you can find the value of x that corresponds to the maximum point on the area curve.
1. Take the derivative of the area with respect to x:
dA/dx = 60 - 4x.
2. Set the derivative equal to zero and solve for x:
60 - 4x = 0
4x = 60
x = 15.
So, the width of the rectangle should be 15 feet.
Substitute this value of x back into the area formula to find the length:
Length = 60 - 2x = 60 - 2(15) = 60 - 30 = 30 feet.
Therefore, the dimensions of the rectangle that will give you the largest possible area for the garden are 15 feet by 30 feet.
To find the dimensions of the rectangle that will result in the largest possible area, you can use calculus to maximize the area function. However, in this case, we can solve it using basic algebra and optimization.
Let's denote the length of one of the sides of the rectangle as 'x', which is perpendicular to the side of the house. The two other sides parallel to the house will each have length 'y'. We are given that the total fence length is 60 feet, so the equation for the perimeter of the rectangle is:
2x + y + y = 60
2x + 2y = 60
x + y = 30
Now, we need to express the area of the rectangle in terms of x and y. The area of a rectangle is given by A = x * y.
Since we have an equation for x + y, we can solve for y in terms of x: y = 30 - x.
Substituting this into the area function: A = x * (30 - x).
To maximize A, we need to find the value of x that corresponds to the maximum area. To do this, we can express the area as a quadratic function (f(x) = -x^2 + 30x) and find the vertex of the parabola.
The formula to find the x-coordinate of the vertex of a quadratic function in the form f(x) = ax^2 + bx + c is x = -b / (2a).
In our case, a = -1, b = 30, and c = 0. Plugging these values into the formula gives us:
x = -30 / (2 * -1)
x = -30 / -2
x = 15
So, the x-coordinate of the vertex is 15. Since the length of the garden cannot be negative, we know that x = 15 feet.
To find the corresponding y-coordinate, we can substitute this value into the equation we found earlier: y = 30 - x = 30 - 15 = 15.
Therefore, the dimensions of the rectangle that will result in the largest possible area are 15 feet by 15 feet.