How should this be done?
Suppose you have 132 m of fencing with which to make two side-by-side rectangular enclosures against an existing wall. if the rectangular enclosures are adjacent and of the same depth what is the maximum are that can be enclosed?
thanks!
To find the maximum area that can be enclosed, we can set up an equation based on the perimeter of the rectangular enclosures.
Let's assume the width of each enclosure is x meters. Since we have two enclosures side by side, the total length of fencing used for the widths would be 2x meters.
We also know that the depth of each enclosure is the same. Let's assume the depth is d meters.
The total length of fencing used for the depths would be 2d meters.
Given that we have 132 meters of fencing in total, the equation would be:
2x + 2d = 132
To maximize the area, we want to express the area in terms of a single variable. Since the question states that the depths of the enclosures are the same, we can write the area of one enclosure as length times width, which is d * x.
To maximize the area, we need to solve for d and x in terms of each other. We can do this by expressing d in terms of x from the perimeter equation:
2x + 2d = 132
2d = 132 - 2x
d = (132 - 2x)/2
d = 66 - x
Now we can substitute this value of d into the area equation:
Area = d * x
Area = (66 - x) * x
To find the maximum area, we need to find the vertex of the quadratic equation representing the area. This can be done by finding the x-coordinate of the vertex using the formula:
x = -b/2a
For our equation (with a = -1, b = 66, and c = 0), we get:
x = -66/2(-1)
x = 66/2
x = 33
Substituting this value of x back into the equation for the depth, we get:
d = 66 - x
d = 66 - 33
d = 33
Therefore, the maximum area that can be enclosed is when the width (x) is 33 meters and the depth (d) is also 33 meters. The maximum area would be:
Area = d * x
Area = 33 * 33
Area = 1089 square meters
So, the maximum area that can be enclosed is 1089 square meters.
To find the maximum area that can be enclosed, we need to break down the problem into smaller steps:
1. Understand the problem: We have 132 meters of fencing to make two rectangular enclosures against an existing wall. The enclosures are adjacent (side-by-side) and have the same depth.
2. Identify the variables: Let's call the length of each enclosure "x" and the depth "y".
3. Set up constraints and equations: The total amount of fencing we have is 132 meters, which means the sum of the perimeters of the two enclosures should equal 132 meters. The perimeter of a rectangle is given by the formula P = 2(x + y).
Therefore, 2x + 4y = 132, as we have two rectangular enclosures.
4. Simplify the equation: Divide both sides of the equation by 2 to obtain x + 2y = 66.
5. Solve for x in terms of y: Rearrange the equation to solve for x: x = 66 - 2y.
6. Find the area: The area of a rectangle is given by the formula A = length × width, which means the area of each enclosure is xy.
Substitute the value of x we found in step 5: A = (66 - 2y)y.
7. Maximize the area: To find the maximum area, we need to find the maximum value of the function A = (66 - 2y)y. This can be done by finding the vertex of a quadratic function.
The vertex of the quadratic function A = ay^2 + by + c is given by the coordinates (-b/2a, A).
In our case, a = -2, b = 66, and c = 0. Substituting these values into the formula, we find that the vertex is (-b/2a, A) = (-66/-4, A) = (33, 1089).
Therefore, the maximum area that can be enclosed is 1089 square meters.
Final answer: The maximum area that can be enclosed in two side-by-side rectangular enclosures is 1089 square meters.
Let the combined length of the two rectangles by y
let the width be x (there will be 3 of those)
so 3x + y = 132
y = 132-3x
Area = xy
= x(132-x)
= -x^2 + 132x
by Calculus
d(Area)/dx = -2x + 132 = 0 for a max of Area
x = 66
then max Area = 66(132-66) = 4356 m^2
by completing the square:
Area = -[x^2 - 132x + 4356 - 4356 ]
= -(x-66)^2 + 4356
so the max Area is 4356 , when x = 66