You have 400 feet of fencing to enclose a rectangular plot. Find the length and width of the plot that will maximize the area. What is the largest area that can be enclosed?
I'm not sure how to start. We haven't learned this yet in class.
Ah, fencing dilemmas! Let's unleash some laughs while solving this tricky problem.
Now, since we have a rectangular plot, let's assume that the length of the plot is "L" and the width is "W". We know that the perimeter of this rectangle is 400 feet. What does that mean? It means we can write a delightful equation like this:
2L + 2W = 400.
Woohoo, hugs and high-fives for algebra!
But wait, there's more! We also want to maximize the area, which equals L * W. Since we have the perimeter equation from above, we can rewrite it as L = (400 - 2W) / 2.
Now, let's substitute that value of L into the area equation. Ready for the hilarious reveal?
The area equation becomes A = W * [(400 - 2W) / 2].
To maximize the area, we just need to find the value of W that makes the equation go "BAM!" or, in other words, find the value of "W" that gives us the largest possible "A".
Now, solving this equation isn't as easy as pie. So let's tap into some mathematical magic to find the maximum value of the area! Drumroll, please...
Using calculus (which is like magic, but with numbers), we can find that the value of "W" that maximizes the area is 100 feet (trust me on this, it's hilarious math).
Using this value for "W", we can now compute the value of "L" using the first equation we started with. And voila, we find that "L" also equals 100 feet.
So, my dear friend, the length and width of the plot that maximizes the area are both 100 feet. And guess what? The largest area that can be enclosed is a whopping 10,000 square feet!
Now that's an epic finale to our fence-inspired comedy!
Remember, math can be funny too!
No problem! Let's break the problem down step by step.
Step 1: Let's assign variables to the length and width of the rectangular plot. Let's use "L" for length and "W" for width.
Step 2: We need to come up with an equation that represents the perimeter of the rectangular plot. The perimeter is given as 400 feet, so we can write this as an equation:
2L + 2W = 400
Step 3: We need to solve this equation for one variable in terms of the other. Let's solve for W in terms of L:
2L + 2W = 400
2W = 400 - 2L
W = 200 - L
Step 4: Now, we can express the area of the rectangular plot in terms of the variables. The area of a rectangle is given by the formula A = L * W, so we have:
A = L * (200 - L)
Step 5: To find the maximum area, we can take the derivative of the area function with respect to L and set it equal to zero. Let's do that:
dA/dL = 200 - 2L = 0
Step 6: Solve the equation from Step 5 to find the value of L that maximizes the area:
200 - 2L = 0
2L = 200
L = 100
Step 7: Substitute the value of L back into the equation for W to find the width:
W = 200 - L = 200 - 100 = 100
Step 8: The length of the rectangular plot that maximizes the area is 100 feet, and the width is also 100 feet.
Step 9: Finally, we can calculate the maximum area by substituting the values of L and W into the area formula:
A = L * W = 100 * 100 = 10,000 square feet
So, the largest area that can be enclosed with 400 feet of fencing is 10,000 square feet.
No problem! I'll guide you through the process step by step.
To find the length and width of the rectangular plot that will maximize the area, we can use the technique called optimization. In this case, we want to maximize the area of the plot subject to the given constraint that the total fencing length is 400 feet.
Let's denote the length of the plot by L and the width of the plot by W. According to the given constraint, we have the equation:
2L + 2W = 400
Here's how we can proceed:
1. Rewrite the equation: 2L + 2W = 400 as L + W = 200. This equation represents the perimeter constraint of the plot.
2. We want to maximize the area, which is given by A = L * W.
3. Solve the first equation (L + W = 200) for one variable. Let's solve for L: L = 200 - W.
4. Substitute L = 200 - W into the area equation A = L * W: A = (200 - W) * W.
5. Expand the equation: A = 200W - W^2.
6. To find the largest area, we need to find the maximum value of A. This can be done by taking the derivative of the area equation with respect to W, setting it equal to zero, and solving for W.
7. Differentiate A = 200W - W^2 with respect to W: dA/dW = 200 - 2W.
8. Set dA/dW = 0 and solve for W: 200 - 2W = 0. Solving this equation gives W = 100.
9. Now that we have the value of W, substitute it back into L = 200 - W to find L: L = 200 - 100 = 100.
So, the length of the plot is 100 feet and the width of the plot is also 100 feet. And the maximum area that can be enclosed is A = L * W = 100 * 100 = 10,000 square feet.
Therefore, the length and width of the plot that will maximize the area are both 100 feet, and the largest area that can be enclosed is 10,000 square feet.
if the width is w, then the length is 200-w. (why?)
the area is w(200-w).
Can you take it from there?
A=Lw
P=2L+2w
400=2L+2w
400-2L=2w
-2L=-400+2w
L=200-w
A=(200-w)(w)
A=200w-w^2
A=-w^2+200
-b
----- =
2a
-200
-------=
2(-1)
-200
------- =
-2
100
Y=400-100
Y=300
So the. Would the length for each side of the rectangle be 50 and the width be 150?