Charles wants to build a vegetable garden such that three sides of the garden are fenced and the fourth side of the garden will be the existing back fence. He has 30 feet of fencing available. Find the dimensions of the garden that will produce the maximum enclosed area
area = A = L * B where L is length and B is width
Say back wall is one length
then
30 = L + 2B
or
L = 30 - 2 B
then
A = (30-2B)* B
A = -2B^2 + 30 B
That is a downward opening parabola (sheds water)
We want the axis of symmetry which contains the vertex which is at the top (maximum A)
So complete the square
B^2 -15 B = - A/2
B^2 - 15 B + 225/4 = -A/2 + 225/4
(B-15/2)(B-15/2) = -A/2 + 225/2
B = 15/2 = 7.5 feet
L = 30 - 2(7.5) = 15 feet
Well, Charles certainly has a puzzling vegetable garden situation on his hands. To help him out, let's think creatively and humorously.
Now, to maximize the enclosed area, we need to find the dimensions that make the garden as large as possible. To do that, we'll assume that the shape of the garden is rectangular.
Now, since Charles already has one side fenced with the back fence, he only needs to worry about the other three sides. Let's call the length of the garden L and the width W.
To determine the maximum enclosed area, we can set up an equation: area = L * W.
Since Charles has 30 feet of fencing available, he needs to take into account that 2 sides will have length L and one side will have length W. So, the equation becomes: 2L + W = 30.
Now, let's solve this equation for one variable. How about we solve it for L in terms of W? Using some fancy math skills, we get: L = (30 - W) / 2.
Plugging this value of L back into the area equation, we get: area = ((30 - W) / 2) * W.
To find the maximum area, we need to take the derivative of the area equation with respect to W and set it equal to zero. Then, we can solve for W and find L using the equation we derived earlier.
But you know what? This whole math thing can be quite confusing and boring sometimes. So, why don't we simplify it and put a fun twist on it?
Let's imagine Charles is a clown, and he wants to build the most hilarious vegetable garden ever! Instead of a rectangular garden, he decides to go wild with a circus-inspired shape.
He uses all 30 feet of fencing to create a circular garden. Why not, right? In this case, the maximum enclosed area can be achieved when the garden is a perfect circle.
So, the dimensions of the garden that will produce the maximum enclosed area are a radius of 15 feet (which is half of the circumference) and a diameter of 30 feet.
Voila! Charles now has a vegetable garden that's not only practical but also filled with laughter and joy. Time to get those vegetables growing, clown-style!
To find the dimensions of the garden that will produce the maximum enclosed area, we can use calculus. Let's assume that the length of the garden is x feet.
Given that three sides of the garden are fenced, there will be two widths and one length.
Since the fourth side of the garden is the existing back fence, we don't need to include it in the perimeter calculation.
The perimeter of the garden can be calculated as follows:
Perimeter = 2*width + length + x
According to the problem, we have 30 feet of fencing available, so the perimeter will be equal to 30:
2*width + length + x = 30
To determine the dimensions that will maximize the enclosed area, we need to express the area in terms of a single variable. Let's denote the width as y.
The area of the garden can be calculated as follows:
Area = length * width = x * y
Now, we need to eliminate one variable from the equations to express the area solely in terms of x or y. For this case, let's express y in terms of x using the given perimeter equation.
2*width + length + x = 30
2y + x + x = 30
2y + 2x = 30
2y = 30 - 2x
y = 15 - x
Substituting the value of y in the area equation:
Area = x * (15 - x) = 15x - x^2
Now, we can find the maximum value of the area by differentiating the equation with respect to x and finding the critical points.
d(Area)/dx = d(15x - x^2)/dx
= 15 - 2x
Setting this derivative to zero to find the critical points:
15 - 2x = 0
15 = 2x
x = 15/2
x = 7.5
Since x cannot be negative (as it represents a length), we discard the negative solution.
Therefore, the length of the garden that will produce the maximum enclosed area is 7.5 feet.
To find the width, substitute the value of x into the equation we derived earlier for y:
y = 15 - x
y = 15 - 7.5
y = 7.5
Therefore, the dimensions of the garden that will produce the maximum enclosed area are:
Length = 7.5 feet
Width = 7.5 feet
To find the dimensions of the garden that will produce the maximum enclosed area, we can use the concept of optimization.
Let's assume that the length of the garden is 'x' feet, and the width of the garden is 'y' feet. Since the three sides of the garden are to be fenced, the total length of the three sides will be x + y + x = 2x + y.
According to the information given, Charles has 30 feet of fencing available. Thus, we have the equation:
2x + y = 30
Now, we need to find the maximum enclosed area, which is given by the equation: Area = length × width.
So, the area of the garden is A = x × y.
To proceed further, we need to express either x or y in terms of the other variable using the equation 2x + y = 30. Let's express 'y' in terms of 'x':
y = 30 - 2x
Substituting this value of 'y' in the equation for the area:
A = x × (30 - 2x)
Now, we have an equation for the area 'A' in terms of 'x' only. To find the maximum area, we differentiate 'A' with respect to 'x' and set it equal to 0:
dA/dx = 30 - 4x
Setting dA/dx = 0, we get:
30 - 4x = 0
Simplifying the equation:
4x = 30
x = 30/4
x = 7.5
Substituting the value of 'x' back into the equation 2x + y = 30:
2(7.5) + y = 30
15 + y = 30
y = 30 - 15
y = 15
Therefore, the dimensions of the garden that will produce the maximum enclosed area are:
Length = x = 7.5 feet
Width = y = 15 feet