A farmer wants to make three identical rectangular enclosures along a straight river, as in the diagram shown below. If he has 1200 yards of fence, and if the sides along the river need no fence, what should be the dimensions of each enclosure if the total area is to be maximized?
Let,
x = length
y = width
We know that the total length fences is 1200 m
l 3x + y = 1200
l y = 1200-3x (1)
l A = xy (2)
Substitute (1) in (2)
l A = x(1200-3x)
l A = 1200x-3x2
Domain: 0 ≤ x ≤
Using derivatives,
l = 1200-6x
Optimum: l 1200-6x=0
l 6x = 1200
Answer:
x = 200m
y = 1200 - 3(200)
y = 600m
To maximize the total area of the three identical rectangular enclosures, we need to find the dimensions that will result in the largest area.
Let's denote the width of each enclosure as "w" and the length as "L."
Given the constraints, we know that the three enclosures will share two side fences, which are parallel to the river. This means that we only need to consider the fences separating the enclosures.
Since there are three enclosures, we will need two fences to separate them. This leaves us with the remaining fence length of 1200 - 2L.
To maximize the area, we need to find the dimensions that will give us the largest possible area for the total three enclosures.
The area of a rectangular enclosure is given by the formula: A = width * length.
For the first enclosure, the width will be "w", and the length will be "L". So, the area of the first enclosure (A1) is A1 = w * L.
For the second enclosure, the width will also be "w", and the length will be the remaining length after subtracting the fences: 2L (since there are two fences separating the enclosures). So the area of the second enclosure (A2) is A2 = w * (1200 - 2L) / 2.
For the third enclosure, the width will also be "w", and the length will be the remaining length after subtracting the fences from both sides: L. So the area of the third enclosure (A3) is A3 = w * L.
The total area (Atotal) of the three enclosures is the sum of A1, A2, and A3: Atotal = A1 + A2 + A3.
Now, we can express Atotal as a function of L:
Atotal = w * L + w * (1200 - 2L) / 2 + w * L
Atotal = 2wL + w(1200 - 2L) / 2
Atotal = 2wL + w(600 - L)
Atotal = 2wL + 600w - wL
Atotal = wL + 600w
To find the values of w and L that will maximize the area, we will need to find the derivative of Atotal with respect to L and set it equal to zero.
d(Atotal) / dL = w + 600 = 0
Solving for w, we get w = -600.
However, since we are looking for positive dimensions, we can ignore the negative value. Therefore, w = 600.
Now, substitute the value of w back into the equation for Atotal to solve for L:
Atotal = wL + 600w
Atotal = 600L + 600(600)
Atotal = 600L + 360000
Since the length along the river does not require a fence, we can assume L < 600.
Taking the derivative of Atotal with respect to L:
d(Atotal) / dL = 600
Setting this derivative equal to zero:
600 = 0
Since there is no solution to this equation, we can conclude that the maximum area is obtained when L approaches the upper bound of L < 600.
Therefore, to maximize the total area of the three identical rectangular enclosures, each enclosure should have a width (w) of 600 yards, and the length (L) can be any value less than 600 yards.
To maximize the total area of the three enclosures, we need to find the dimensions of each enclosure that will give us the maximum area.
Let's denote the length of each enclosure as L and the width as W. Since there are three identical enclosures, we need to find the dimensions that will maximize the total area of three rectangles.
The perimeter of each enclosure (excluding the sides along the river) can be calculated using the given information that there are 1200 yards of fence. Since there are two lengths and two widths of each enclosure, the equation becomes:
2L + 2W = 1200
We need to express one of the variables in terms of another so that we can rewrite the area equation in a single variable.
Let's solve the equation for L:
2L = 1200 - 2W
L = (1200 - 2W) / 2
L = 600 - W
Now, we can express the total area of three enclosures in terms of W only:
Total Area, A = 3LW
A = 3(600 - W)W
A = 1800W - 3W²
To maximize the area, we need to find the critical points of the area function by taking the derivative and setting it equal to zero:
dA/dW = 1800 - 6W
Setting dA/dW equal to zero:
1800 - 6W = 0
6W = 1800
W = 300
Now that we have the width as 300, we can substitute this value back into the equation for L:
L = 600 - W
L = 600 - 300
L = 300
Therefore, the dimensions of each enclosure should be 300 yards by 300 yards to maximize the total area.
Let the length be 3x, and the width be y.
Maximum area is achieved when 3x = 4y = 600
So the dimensions are 600 by 150
To see this, note that
3x + 4y = 1200
The area is
a = 3xy = 3x(1200-3x)/4
find that da/dx = 0 when x = 200
In all these area problems, maximum area is when the available fencing is divided equally among the lengths and the widths.