You have 8L feet of fence to make a rectangular vegetable garden alongside the wall of your house, where L is a positive constant. The wall of the house bounds one side of the vegetable garden. What is the largest possible area for the vegetable garden?
Well, well, well! Looks like you're trying to maximize your veggie garden area. I'm on it like a carrot on a stick!
Let's break it down, shall we? You've got 8L feet of fence and one side is already covered by the house wall. That means you've got three sides to fence in, right?
Since we're going for a rectangular garden, let's call the length of the garden L (keeping things consistent) and the width W. The total fence length you have to play with is 2L + W (two sides of length L and one side of width W).
But hold up! There's a catch. The total fence length you have, 2L + W, should equal the given 8L feet of fence. So, we have our first equation: 2L + W = 8L.
Now, let's move on to the area, my veggie-loving friend! The area of a rectangle is simply length times width, right? So, the area of our garden is L x W.
Now, remember what we want to do? We want to maximize the area! In other words, find the largest possible value for L x W.
Here's a trick. Remember our first equation? Solve it for W. We get W = 8L - 2L = 6L.
Now, substitute W = 6L into our area equation to get L x 6L. Simplify this to 6L^2, and voila! We've got our area equation in terms of a single variable.
To find the largest possible area, we can take the derivative of the area equation, set it equal to zero, and solve for L. But hey, let's not get too mathy here!
The key takeaway is that the largest possible area for your vegetable garden is 6L^2. So, my veggie-loving friend, go forth and plant those veggies in a garden with maximum clown-approved area! Happy growing!
To find the largest possible area for the vegetable garden, we need to determine the dimensions of the rectangle that will maximize the area.
Let's assume the length of the vegetable garden is L and the width is W. Since the wall of the house bounds one side of the garden, we can say that W + L + W = 8L (because we have 8L feet of fence available). Simplifying this equation, we get:
2W + L = 8L
Now, we want to express W in terms of L, so we can substitute it back into the equation of area later. Rearranging the equation, we get:
2W = 8L - L
2W = 7L
W = (7L)/2
Now that we have expressed W in terms of L, we can find the area of the rectangle (A) by multiplying the length by the width:
A = L * W
A = L * (7L)/2
A = (7L^2)/2
So, the area of the vegetable garden is given by (7L^2)/2.
To find the largest possible area, we need to maximize this expression. Since L is a positive constant, we can see that the area will be largest when L is largest. Therefore, the largest possible area for the vegetable garden is when L is at its highest value, which means the length is the maximum value allowed by the amount of fence available.
Since we have 8L feet of fence, the maximum value for L is 8L/4 = 2L. Therefore, the largest possible area for the vegetable garden is:
A = (7(2L)^2)/2
A = (7(4L^2))/2
A = 14L^2
Hence, the largest possible area for the vegetable garden is 14L^2.
To find the largest possible area for the vegetable garden, we need to understand the constraints and formulate an equation. Here's how you can do it:
1. Let's assume the width of the garden is 'w' and the length is 'L'. The width will be perpendicular to the house wall, and the length will be parallel to it.
2. Since the house wall bounds one side of the garden, the sum of the widths (w) on the other side of the garden should be equal to the remaining length of the fence, which is (8L - w).
3. The perimeter of the rectangle is given by: 2w + L + (8L - w) = 9L
4. Since the perimeter is given, we can isolate the width in terms of L: 2w + L = 9L - w
5. Solving this equation, we find that w = 4L
6. The area of the rectangle is given by: A = length × width = L × w = L × 4L = 4L^2
7. Now, we need to find the value of L that maximizes the area. To do this, we can take the derivative of the area equation with respect to L and set it equal to zero.
8. Differentiating 4L^2, we get: dA/dL = 8L
9. Setting 8L equal to zero, we find L = 0.
10. Since L is a positive constant, we discard L = 0.
11. Therefore, the largest possible area for the vegetable garden is obtained at L = 0.
12. Substituting L = 0 into the area equation, we find A = 4(0)^2 = 0.
Hence, the largest possible area for the vegetable garden is 0 square units.
Divide the fence into two equal parts.
Allocate each half to the number of lengths or widths.
8L/2 = 4L
So, assuming the house is along the long side, there is one length and two widths.
Each length is 4L
Each width is 4L/2 = 2L
Thus the maximum area is 4L*2L = 8L^2