Vitaly and Jen have 24m of fencing to enclose a vegetable garden at the back of their house. What are the dimensions of the largest rectangular garden they could enclose with this length of fencing?

yeah, its 3 sides thanks

I need help for homework plz, Sam brought 24m of fencing to put around his new garden. What could the dimensions of his garden be? What would the area be.

Well, let's start by trying to think inside the box, or rather, the rectangle! If we want to maximize the area of a rectangular garden, we need to find the dimen-sion that can give us the largest possible area. So, let's break out the math tools and solve this green mystery!

Since they have 24 meters of fencing, we can assume that they can enclose all four sides of the rectangular garden. In that case, we have two pairs of equal sides, let's call them 'x' and 'y.'

To maximize the area, we need to find the right values for 'x' and 'y' that satisfy the equation: 2x + 2y = 24. This equation represents the total amount of fencing used.

We can simplify this equation to x + y = 12. And now it’s time to bring in the big guns: humor! Let's say 'x' is the amount of vegetables Jen and Vitaly farm together, and 'y' is the number of jokes I crack while lurking in the garden. So, we want to find the optimal combination of veggies and jokes, which is equal to 12.

Now, we need to remember that we are looking for the largest rectangular garden, so we want to maximize the area, A, which is equal to x * y.

Since we have the equation x + y = 12, we can solve it for y by subtracting 'x' from both sides, which gives us y = 12 - x.

Now we can substitute this value of y back into the area equation: A = x * (12 - x).

To maximize the area, we need to take the derivative of A with respect to x, set it equal to zero, and solve for 'x'. But hey, let's take a break from all that math madness and giggle about vegetable jokes until we find the right answer!

To find the dimensions of the largest rectangular garden that can be enclosed with a given length of fencing, we need to apply the concept of optimization. In this case, we want to maximize the area of the rectangular garden.

Let's assume the length of the rectangular garden is L and the width is W (both in meters). Since we have a total of 24 meters of fencing, we can set up an equation that relates the perimeter of the rectangular garden to its dimensions:

2L + 2W = 24

Simplifying this equation further, we get:

L + W = 12

Now, we can solve this equation to express one variable in terms of the other. Let's express L in terms of W:

L = 12 - W

Next, we can formulate the area of the rectangular garden in terms of these variables:

Area = Length × Width
Area = (12 - W) × W
Area = 12W - W^2

The goal is to find the value of W that maximizes the area (Area). To do this, we can take the derivative of the area equation with respect to W and set it equal to zero:

d(Area)/dW = 12 - 2W = 0

Solving for W, we find:

W = 6

Now, substitute this value of W back into the equation for L to find the length:

L = 12 - W
L = 12 - 6
L = 6

Therefore, the dimensions of the largest rectangular garden that can be enclosed with 24 meters of fencing are 6 meters by 6 meters.

Is the fencing on all sides,or just three?

If three sides:
24=2L+w
Area=Lw=l(24-2L)

Using calculus, set the derivative to zero.
0=24-2L -2L
solve for L.Then go back and solve for W from the perimeter.