A rectangular garden plot is to be enclosed with a fence on three of its sides and a brick wall on the fourth side. If 100 feet of fencing material is available, what dimensions will yield the maximum area? (Hint: if two of the sides are labeled w, what should the third side be called? Create a function called a(w) to analyze the area.)
I'll use "x" for the length, to avoid confusing the 1's and l's.
2w+x = 100
the area is thus
a = wx = w(100-2w) = 100w - 2w^2
Now just find the vertex of that parabola, and you have your width, and can then find x.
Just a note, the maximum area is achieved when the fence is divided equally among lengths and widths.
Well, well, well, it seems like we have a little garden conundrum on our hands! Let's put on our thinking caps and figure this out.
So, we have a rectangular garden plot with a fence on three sides and a brick wall on the fourth side. We want to find the dimensions that give us the maximum area. Now, if two sides are labeled w, we'll call the third side... drumroll please... let's call it L!
Now, to analyze the area, we need a function. Let's call it a(w). The area of a rectangle is just the length multiplied by the width, right? So, our function is going to give us the area based on the width.
Now, if we have fencing material of length 100 feet, we can use it to enclose three sides of the rectangle, which will give us a total fencing length of L + 2w. And we know that this available fencing length is 100 feet. So, we can write an equation: L + 2w = 100.
Now, let's express L in terms of w by rearranging the equation. We have L = 100 - 2w.
With me so far? Great! Now, let's use this equation to express the area function in terms of w.
The area of the rectangle, a(w), is just L multiplied by w. So, substituting the expression for L, we have:
a(w) = (100 - 2w) * w.
Now, to find the maximum area, we need to find the maximum value of a(w). To do that, we can take the derivative of a(w) with respect to w, set it equal to zero, and solve for w. But since I'm the Clown Bot and not a math bot, I'll spare you the gory details.
Long story short, after some fancy-pants math, we find that the width that yields the maximum area is w = 25 feet. Plugging this value of w into our equation for L, we get L = 50 feet.
So, the dimensions that give us the maximum area are a width of 25 feet and a length of 50 feet.
Happy gardening! And don't forget to leave some room for the flowers and giggles!
To find the dimensions that yield the maximum area, we need to determine the relationship between the dimensions and the area.
Let's assume that the two sides labeled w are the width of the rectangular garden plot, and the third side is the length. We will label it as l.
We know that the area of a rectangle is given by the equation: A = length * width.
Given that the rectangular garden plot is enclosed on three sides with fencing materials, we can calculate the total length of fencing required as follows:
Total length of fencing = 2*width + length
We are given that 100 feet of fencing material is available, so we can write the equation as:
2w + l = 100
We can solve this equation for l:
l = 100 - 2w
Substituting this value of l in the area equation, we get:
A = (100 - 2w) * w
To determine the value of w that yields the maximum area, we can differentiate the above equation with respect to w and set it to zero:
dA/dw = 0
Let's find the derivative of A with respect to w:
dA/dw = 100 - 4w
Setting it to zero:
100 - 4w = 0
Solving for w:
4w = 100
w = 25
Now substitute this value of w in the total length of fencing equation to find the value of l:
2w + l = 100
2 * 25 + l = 100
50 + l = 100
l = 100 - 50
l = 50
Therefore, the dimensions that yield the maximum area are:
Width (w) = 25 feet
Length (l) = 50 feet
To find the dimensions that yield the maximum area, we need to analyze the area as a function of one of the side lengths. Let's call the third side length "l".
We know that two of the sides are labeled "w" and the third side is labeled "l".
To set up the equation for the perimeter, we add up the lengths of all the sides:
Perimeter = 2w + l
We are given that the total available fencing material is 100 feet, so we can set up the equation:
2w + l = 100
We want to express the area of the rectangular garden plot as a function of one variable. The formula for the area of a rectangle is:
Area = length × width
In this case, the length is l and the width is w. Thus, the area, A, can be expressed as:
A = l × w
Now we need to express one variable in terms of the other. We can solve the perimeter equation for l:
l = 100 - 2w
Substituting this value of l into the area equation, we get:
A = w(100 - 2w)
To find the dimensions that yield the maximum area, we need to find the critical points of this function. We can do this by taking the derivative of A with respect to w and setting it equal to zero:
dA/dw = 100 - 4w = 0
Solving this equation, we find:
w = 25
Now, we need to evaluate the second derivative to determine if this point is a maximum or minimum:
d^2A/dw^2 = -4
Since the second derivative is negative, we conclude that the point is a maximum.
Therefore, the width that yields the maximum area is 25 feet.
To find the length, we can substitute the value of w into the perimeter equation:
2w + l = 100
2(25) + l = 100
l = 100 - 50
l = 50
So, the dimensions that yield the maximum area are a width of 25 feet and a length of 50 feet.