Diana has 520 yards of fencing to enclose a rectangular area. Find the dimensions of the rectangle that maximize the enclosed area. What is the maximum area?
It would be a square measuring 130 yards on each side.
Oh, fencing and rectangles, how exciting! It's time for "Clown Bot" to crunch some numbers and bring some humor to the party.
Let's imagine Diana as a fencing enthusiast, trying to maximize her rectangle of fun! Now, once upon a time, Diana had 520 yards of fencing...
I bet she really got herself in quite the "loop" with all that fencing, right? 🤡
Now, we know that the perimeter of a rectangle is the sum of all the sides. Since Diana has 520 yards of fencing, we can set up an equation like this:
2L + 2W = 520
Let's move on to the next act! 🎪
To maximize the enclosed area, we need to express the area as a function of one variable, so we can find the maximum point. Let's use the formula:
A = L * W
Now, let's solve this equation in terms of one variable. We can use the perimeter formula to find L (length) in terms of W (width):
L = (520 - 2W) / 2
Plugging that back into the area formula, we have:
A = W * [(520 - 2W) / 2]
Now, let's make that equation a bit more whimsical and expand it:
A = 260W - W^2
Oh, look at that! We've got ourselves a quadratic equation! And quadratic equations mean maxima and minima. Let's find the maximum point, shall we? 🕺
To find the maximum point, we need to take the derivative of the area equation with respect to W, and set it equal to zero.
dA/dW = 260 - 2W
Let's set this equal to zero and solve:
260 - 2W = 0
2W = 260
W = 130
We're halfway there, folks! Now, let's plug this value of W back into our area equation to find the length (L):
L = (520 - 2W) / 2
L = (520 - 2 * 130) / 2
L = 260 / 2
L = 130
Congratulations! We've found the dimensions of the rectangle that maximize the enclosed area. The width is 130 yards, and the length is also 130 yards.
Now, let's find the maximum area by plugging these values into our original area equation:
A = L * W
A = 130 * 130
A = 16900
Ta-da! The maximum enclosed area is 16900 square yards. Diana can now enjoy a ginormous rectangle, perfect for all her sumptuous fencing activities! Huzzah! 🎉
To find the dimensions of the rectangle that maximize the enclosed area, we can use calculus.
Let's assume the length of the rectangle is "l" and the width is "w".
According to the problem, the perimeter of the rectangle is 520 yards:
2l + 2w = 520
Dividing both sides by 2:
l + w = 260
Now, we can express one variable in terms of the other. Let's solve the equation for "w":
w = 260 - l
The area of a rectangle is given by the formula:
A = l * w
Substituting the equation for "w" into the area formula:
A = l * (260 - l)
Now, we have the area A as a function of l. To find the maximum area, we need to find the critical points by taking the derivative of A with respect to l and setting it equal to zero:
dA/dl = 260 - 2l
Setting the derivative equal to zero:
260 - 2l = 0
Solving for l:
2l = 260
l = 130
Now, substitute the value of l into the equation for "w":
w = 260 - l
w = 260 - 130
w = 130
So, the dimensions of the rectangle that maximize the enclosed area are l = 130 and w = 130.
To find the maximum area, substitute the values back into the area formula:
A = l * w
A = 130 * 130
A = 16900 square yards
Therefore, the maximum area is 16900 square yards.
To find the dimensions of the rectangle that maximize the enclosed area, we'll use the concept of optimization.
Let's assume the length of the rectangle is "L" and the width is "W." The perimeter of the rectangle is given as 520 yards, so we can write the equation:
2L + 2W = 520
Simplifying the equation, we get:
L + W = 260
Now, the enclosed area is given by the formula A = L * W.
To maximize the area, we can use a technique called differentiation. We need to find the critical points of the area function and determine whether they correspond to a maximum.
Differentiating the area function with respect to either L or W, we get:
dA/dL = W
dA/dW = L
Setting both derivatives to zero, we have:
W = 0
L = 0
From the perimeter equation (L + W = 260), we can see that both L and W cannot be zero, so this case is not valid.
Next, we'll check the endpoints of the interval. Since we're dealing with lengths, we'll consider non-negative values only.
When L = 0, W = 260.
When W = 0, L = 260.
So, the valid dimensions for the rectangle are (0, 260) or (260, 0).
Now, let's calculate the corresponding areas for these two cases:
Case 1: L = 0, W = 260
A1 = 0 * 260 = 0
Case 2: W = 0, L = 260
A2 = 260 * 0 = 0
In both cases, the area is zero, which suggests that these dimensions are not the maximum.
Since there are no valid critical points or endpoints that maximize the area, it implies that the maximum area occurs at a critical point. In this case, it's at the midpoint L = W = 260 / 2 = 130.
So, the dimensions that maximize the area are L = 130 yards and W = 130 yards, creating a square. The maximum area is given by:
A_max = L * W = 130 * 130 = 16900 square yards.