A homeowner wants to fence a rectangular garden using 60 ft of fencing. An existing stone wall will be
used as one side of the rectangle. Find the dimensions for which the area is a maximum
60 = L + 2 w so L = 60 -2 w
A = L w
A = (60-2 w)w = 60 w - 2 w^2
If you know calculus, just take derivative and set to zero and you are done
If not, complete the square to find vertex of parabola.
w^2 - 30 w = -A/2
w^2 - 30 w + 225 = -A/2 + 225
(w-15)^2 = -(1/2)(A - 450)
so
w = 15
L = 60 - 30 = 30
area = 450 by the way
Nope I still don't get it. I understand right up until this...
"
w^2 - 30 w = -A/2
w^2 - 30 w + 225 = -A/2 + 225
(w-15)^2 = -(1/2)(A - 450)
so
w = 15
L = 60 - 30 = 30
area = 450 by the way
"
Thanks anyway!
So the maximum area would be 450?
I'm trying to read through your work so I can understand how you got the answer
Why did the tomato turn red?
Because it saw the garden in all its rectangular splendor and couldn't contain its excitement!
Now, let's help that homeowner find the dimensions that will maximize their garden's area.
Let's call the length of the rectangle L and the width W.
Since one side of the garden is already covered by an existing stone wall, we only need to fence the other three sides.
The total amount of fencing needed is given as 60 ft. So, we can write the equation:
2L + W = 60 (since we have to fence two lengths and one width)
Solving for W, we get:
W = 60 - 2L
Now, we need to find an equation for the area of the rectangle, A.
The formula for the area of a rectangle is A = L * W.
Substituting the value of W from above, we have:
A = L * (60 - 2L)
Now, we want to find the dimensions for which the area A is a maximum.
To do that, let's take the derivative of A with respect to L and set it equal to zero:
dA/dL = 60 - 4L
Setting this equal to zero and solving for L, we find:
60 - 4L = 0
4L = 60
L = 60/4
L = 15
So, the length of the rectangle should be 15ft.
Substituting this value back into our equation for W, we find:
W = 60 - 2(15)
W = 60 - 30
W = 30
Therefore, the width of the rectangle should be 30ft.
So, the dimensions that will maximize the garden's area are 15ft by 30ft. Happy gardening!
To find the dimensions for which the area is a maximum, we need to maximize the area of the rectangular garden.
Let's assume the length of the rectangular garden is L and the width is W (excluding the side connected to the stone wall).
We know that the perimeter is 60 ft, so the sum of all sides should be equal to 60 ft. Since one side is attached to the stone wall, we have 2 sides of length L, 1 side of length W, and 1 side of the stone wall.
2L + W = 60 - length of the stone wall (Eq. 1)
Next, we can express the area A in terms of L and W. The area of the rectangular garden is simply the product of its length and width.
A = L * W (Eq. 2)
To maximize A, we need to find the values of L and W that yield the maximum value for A.
To solve this problem, we will use the method of substitution. Let's solve Eq. 1 for L in terms of W:
2L = 60 - length of stone wall - W
L = (60 - length of stone wall - W) / 2 (Eq. 3)
Now we substitute Eq. 3 into Eq. 2:
A = [(60 - length of stone wall - W) / 2] * W
Simplifying:
A = (60W - W^2 - W * length of stone wall) / 2
A = -1/2 * W^2 + (60 - length of stone wall) / 2 * W (Eq. 4)
Now we have expressed the area A in terms of W only. Eq. 4 represents a quadratic equation, and the coefficient of W^2 is negative, which means it is a downward-opening parabola. Therefore, the maximum value of A occurs at the vertex of the parabola.
To find the value of W at the vertex, we can use the formula for the x-coordinate of the vertex:
W = -b / (2a)
In our case, a = -1/2 and b = (60 - length of stone wall) / 2. Substituting these values:
W = -[(60 - length of stone wall) / 2] / (-1/2)
W = (60 - length of stone wall) / 2
To find the corresponding value of L, we substitute W back into Eq. 1:
2L + (60 - length of stone wall) / 2 = 60 - length of stone wall
2L + (60 - length of stone wall) / 2 = 60 - length of stone wall
Simplifying:
2L = 60 - length of stone wall - (60 - length of stone wall) / 2
2L = 60 - length of stone wall - 30 + length of stone wall / 2
2L = 30 + length of stone wall / 2
L = (30 + length of stone wall / 2) / 2
Now we have both the length L and width W in terms of the length of the stone wall. We can substitute different values for the length of the stone wall to find the corresponding dimensions (L and W) that will maximize the area A.