suppose you have 102m of fencing to make two side by side rectangles, making one large rectangle when put together. What is the maxium area that you can enclose?
I assume you wont have a double side on the adjoining sides.
If you do this, you have 102 m of perimeter, to do 3 widths (two ends and center fence), and 2 lengths.
Area= lw
P= 3w+2L or L= 1/2(102-3w)
Area= 1/2 w(102-3w)
This is easy in calc, but I assume you are not there yet. So graph it, notice the max Area, read w. Then solve for l.
In Calculus, the maximum Area occurs when dArea/dw=0
dArea/dw=0=1/2 (102-3w) + 1/2w(-3) solve for w.
Thank you so much for the help
To find the maximum area that can be enclosed with 102m of fencing, we need to determine the dimensions of the rectangles that will maximize the combined area.
Let's assume the length of the large rectangle is L and the width of each smaller rectangle is W. Since the two smaller rectangles are placed side by side, the length of the large rectangle will be 2W.
The perimeter of the large rectangle, P, can be expressed as:
P = 2L + 2W + 2W = 2L + 4W
Given that the total length of the fencing is 102m, we can write the equation:
2L + 4W = 102
Now, we need to express the area of the large rectangle, A, in terms of L and W:
A = L * 2W = 2LW
To find the maximum area, we can solve the equation for A in terms of W and substitute it into the perimeter equation to form a quadratic equation. We can then find the values of W and L that will maximize the area.
Let's start by solving the perimeter equation for L:
2L = 102 - 4W
L = (102 - 4W) / 2
L = 51 - 2W
Substituting this value for L into the equation for A:
A = 2(51 - 2W)W
A = 102W - 4W^2
Now, we have the equation for the area in terms of W. To find the maximum area, we need to differentiate A with respect to W and set it to zero, then solve for W.
dA/dW = 102 - 8W
Setting dA/dW = 0:
102 - 8W = 0
8W = 102
W = 102 / 8
W = 12.75
Since the width cannot be a fraction, we round W down to the nearest whole number, W = 12.
Now, substitute the value of W back into L = 51 - 2W:
L = 51 - 2(12)
L = 51 - 24
L = 27
So, the dimensions of the large rectangle will be L = 27 and W = 12.
Finally, we can calculate the maximum area by substituting these values into the equation for A:
A = 2LW
A = 2(27)(12)
A = 648 square meters
Therefore, the maximum area that can be enclosed with 102m of fencing is 648 square meters.
To find the maximum area that can be enclosed using 102 meters of fencing to create two side-by-side rectangles forming one large rectangle, we need to apply the concept of optimization. Let's break down the problem step by step:
1. Define the problem:
We want to determine the maximum area of a rectangle that can be formed by dividing it into two side-by-side rectangles using a total of 102 meters of fencing.
2. Formulate constraints:
The total length of fencing available is given as 102 meters. This fencing is used to create the perimeter of the large rectangle, as well as to separate the two side-by-side rectangles. We can divide the fencing into three segments: two of equal length that connect the small rectangles to the large rectangle and one that forms the common side between the two small rectangles.
3. Formulate the equation:
Let's represent the dimensions of the large rectangle as:
- Length of the large rectangle: L
- Width of the large rectangle: W
We need to optimize the area, which is given by: Area = Length x Width.
4. Determine the relationship between variables:
The total amount of fencing used will be the sum of perimeters of the large rectangle and the two small rectangles. By using the formula for the perimeter of a rectangle, we can write: 2L + 4W = 102 meters. This equation represents the perimeter constraint.
5. Solve the equation:
Rearranging the equation, we get: 2L = 102 - 4W.
Simplifying further, we find: L = 51 - 2W.
Substituting L = 51 - 2W into the area equation, we have: Area = (51 - 2W) x W.
6. Maximize the area:
To find the maximum area, we can differentiate the area equation with respect to W and set it equal to 0. This will give us the value of W at which the area is maximized.
d(Area)/dW = 0
51 - 4W = 0
Solving for W: W = 51/4 = 12.75.
Now, substitute W = 12.75 into L = 51 - 2W: L = 51 - 2(12.75) = 25.5.
Therefore, the dimensions of the large rectangle are: Length = 25.5 meters and Width = 12.75 meters.
7. Calculate the maximum area:
The maximum area can be calculated by substituting the values of Length and Width into the area equation:
Area = (25.5) x (12.75) = 325.125 square meters.
Hence, the maximum area that can be enclosed using 102 meters of fencing to create two side-by-side rectangles, forming one large rectangle, is 325.125 square meters.