you have 704m of fencing to enclose 3 adjacent plots(not the same size) you must use calculus to fine the dimensions for which the total area is as large as possible. Find the largest possible total area.
What shape are the "plots" ?
129
To find the largest possible total area, we'll use calculus and the concept of optimization. We have 704 meters of fencing to enclose three adjacent plots, each with different sizes. Let's denote the widths of the three plots as x, y, and z.
To start, we need a function that represents the total area. Since the plots are adjacent, the total area can be calculated by adding the areas of each plot. The formula for the area of a rectangle is length times width, so the total area can be written as:
A(x, y, z) = x * w1 + y * w2 + z * w3
Where w1, w2, and w3 represent the widths of each plot.
Now, we need to express the width of each plot in terms of the given total perimeter. Each plot has two vertical sides, and the total length of these sides must equal the total perimeter of 704 meters. So we can write the following equation:
2x + 2y + 2z = 704
Simplifying:
x + y + z = 352
Now, we have an equation that relates the widths of the three plots. To express one width in terms of the others, we'll use algebra. Let's solve for z:
z = 352 - x - y
Now, let's substitute this value for z in the area formula:
A(x, y) = x * w1 + y * w2 + (352 - x - y) * w3
Remember, we are looking for the largest possible total area, which means we need to find the critical points of this function. We can do this by finding the derivative of A with respect to x and y and setting it equal to zero.
∂A/∂x = w1 - w3 = 0 (First partial derivative with respect to x)
∂A/∂y = w2 - w3 = 0 (First partial derivative with respect to y)
Now, we have two equations involving the widths of the plots that are equal to each other. We can use these equations to find the relationship between the widths.
w1 - w3 = w2 - w3
w1 = w2
This indicates that w1 and w2 are equal, meaning the widths of the first two plots are equal. However, this does not provide us with the specific values. To find those, let's substitute the expression for w3 back into the equation:
x = y
Now we have a relationship between the widths of the first two plots.
We can now substitute this relationship into the equation we derived earlier:
A(x) = x * w1 + x * w2 + (352 - 2x) * w3
Since x and y are now equal, we can rewrite this as:
A(x) = 2x * w1 + (352 - 2x) * w3
Now we only have one variable, x. To find the largest possible total area, we need to find the critical points of this function. We'll take its derivative with respect to x and set it equal to zero.
dA/dx = 2w1 - 2w3 = 0
Now we have an equation that relates the widths of the plots. Solving for w1:
w1 = w3
This means that the width of the first plot is equal to the width of the third plot.
Now we can substitute this relationship back into our equation for the total area:
A(x) = 2x * w1 + (352 - 2x) * w1
Simplifying:
A(x) = 2w1 * x + w1 * (352 - 2x)
A(x) = w1 * (2x + 352 - 2x)
A(x) = w1 * 352
This result indicates that the total area is constant and only depends on the value of w1, which represents the width of the first plot.
Therefore, to maximize the total area, we need to make the width of the first plot (w1) as large as possible. However, keep in mind that we also have the equation x + y + z = 352 to satisfy. To optimize the dimensions further, you can assume the size of one plot, calculate the sizes of the other two plots, and then calculate their respective areas. Repeat this process for different assumptions until you find the combination that results in the largest area.
Note: The above calculation assumes that the function A(x, y, z) has a maximum. To rigorously prove that this is the case, further analysis would be required, such as the second partial derivative test or verifying the endpoints of the domain (e.g., x, y, z > 0 and x + y + z = 352).