A rectangular garden is to be enclosed by a fence and divided into 5 regions as shown below. Find the values of L and W for which the total area is as large as possible assuming that only 120 ft. of fencing is available.
Steve, Its a rectangle divide into 5 vertical parts. That's the picture that came with the problem
Steve there is a diagram but i don't know how to get the diagram to post.
To maximize the total area of the rectangular garden, we need to find the values of length (L) and width (W) that satisfy the given constraints.
Let's start by understanding the information given in the problem. We know that the garden is divided into 5 regions, which means there are 4 interior fences separating these regions. Since only 120 ft. of fencing is available, we can deduce that the total length of all the fences is equal to 120 ft.
Now, let's consider how the fence is arranged. The total length of the fence consists of the sum of the lengths of the four interior fences and the two lengths of the outer fence. The two lengths of the outer fence are equal to the width (W) because the garden has two sides of length W.
So, we have the equation: 120 ft = 4L + 2W
Now, let's express the total area of the garden in terms of L and W. The area of the garden can be calculated by multiplying the length (L) by the width (W).
Therefore, the total area (A) is given by the equation: A = L * W
To maximize the area (A), we need to find the values of L and W that satisfy the given constraints and maximize the area function.
We can use the equation 120 ft = 4L + 2W to express one variable (L or W) in terms of the other. Let's solve for L:
4L = 120 ft - 2W
L = (120 ft - 2W) / 4
L = 30 ft - 0.5W
Now, substitute this value of L in the equation for the total area:
A = (30 ft - 0.5W) * W
To find the maximum value of A, we need to find the critical points by taking the derivative of A with respect to W and setting it equal to zero:
dA/dW = 30 ft - W = 0
Solving the equation, we find W = 30 ft.
Substituting this value of W back into the equation for L:
L = 30 ft - 0.5(30 ft)
L = 15 ft
Therefore, the values of L and W that maximize the total area are L = 15 ft and W = 30 ft.
no diagram. How about a description?
I can guarantee that the maximum area is achieved when the fencing is divided equally among lengths and widths.