Suppose you have 168 meters 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?
total length = x
depth = y
fencing needed = x + 3 y = 168
A = x y
A = (168-3y)y = 168 y - 3 y^2
3 y^2 - 168 y = -A
y^2 - 56 y = -A/3
y^2 - 56 y + 784 = -A/3 +784
(y-28)^2 = -(1/3)(A-2352)
so vertex is at A = 2352 square meters
by the way
x = 168 -3y = 84
Let the sides parallel to the given boundary be x and the sides perpendicular to the boundary be y.
Then, 2x + 3y = 168 and A = 2xy.
With y = (168 - 2x)/3
A = (336x - 4x^2)/3
dA/dx = 112 - 8x/3 = 0
336 = 8x making x = 84 and y = 28
The area A = 2352m^2.
Alternatively:
Considering all rectangles with a given perimeter, one side being provided by a straight given boundry, which one encloses the largest area?
Letting P equal the given perimeter and "x" the short side of the rectangle, we can write for the area A = x(P - 2x) = Px - 2x^2.
Taking the first derivitive and setting equal to zero, dA/dx = P - 4x = 0, x becomes P/4.
With x = P/4, we end up with a rectangle with side ratio of 2:1.
.....The short side is P/4.The traditional calculus approach would be as follows.
.....The long side is (P - 2(P/4)) = P/2.
Therefore, it can be unequivicably stated that of all possible rectangles with a given perimeter, one side being a given external boundry, the rectangle with side ratio of 2:1 encloses the maximum area.
To find the maximum area that can be enclosed by two side-by-side rectangular enclosures, we need to maximize the overall area. Let's break down the problem and solve it step by step:
1. Let's denote the length of each enclosure as x meters. Since the two enclosures are side by side, the total length of both enclosures will be 2x. We know that the total length of fencing available is 168 meters, so we can set up the equation: 2x = 168.
2. Now, let's determine the depth of the enclosures. Since the enclosures are adjacent and of the same depth, we can denote the depth as y meters.
3. The maximum area is obtained when the enclosures are right up against the existing wall. This means that one side of each enclosure does not require any extra fencing.
4. Each enclosure has two sides of length y and one side of length x. Therefore, the total amount of fencing required for one enclosure is 2y + x.
5. Since we have two enclosures, the total amount of fencing required will be 2(2y + x) = 4y + 2x.
6. We know that the total length of fencing available is 168 meters, so we can set up another equation: 4y + 2x = 168.
7. We now have a system of two equations:
- 2x = 168
- 4y + 2x = 168
8. Solve the first equation for x:
- 2x = 168
- x = 168/2
- x = 84
9. Substitute the value of x into the second equation:
- 4y + 2(84) = 168
- 4y + 168 = 168
- 4y = 168 - 168
- 4y = 0
- y = 0
10. Since y is equal to 0, this means that the depth of the enclosures is zero. This would imply that the enclosures have no area, which doesn't make sense.
11. Therefore, there is no maximum area that can be enclosed in this scenario.
In summary, given the constraints and the amount of fencing available, it is not possible to maximize the area enclosed by two side-by-side rectangular enclosures.