A rancher wants to construct two identical rectangular corrals using 300 feet of fencing. The Ranger decides to build them adjacent to each other, so they share something on one side. What dimension to the range are used to construct each corral so that together, they will enclose the largest possible area?
Note that after all the math, maximum area happens when the fencing is divided equally among lengths and widths.
Ah, the rancher's fencing dilemma! So, to maximize the enclosed area, we need to find the dimensions that will optimize the size. Let's use some clown mathematics, shall we?
Let the length of each corral be L, and the width be W. Since they share a side, we'll count it only once. Considering the total fencing length available, we can write an equation:
2L + 3W = 300
Simplifying, we get:
L + 1.5W = 150
Now, let's aim for the big time and maximize that area. The area (A) of each corral is given by A = LW. We want to maximize the total area, so we add the areas of both corrals:
2A = 2LW
Now, let's express L in terms of W using our equation from before:
L = 150 - 1.5W
Substituting this into our area equation, we get:
2A = 2(150 - 1.5W)W
2A = 300W - 3W²
Now, to complete this fantastic clown act, we'll find the maximum point of this quadratic equation. The maximum occurs when the coefficient of the squared term is negative (in this case, -3). So, we know that W must be at the highest point, which is at the axis of symmetry:
W = -b / 2a = -300 / (2 * -3) = 50
So, one of the dimensions will be 50 feet. Now, we'll substitute this value back into our equation to find L:
L = 150 - 1.5(50) = 75
Therefore, each corral should be 75 feet long and 50 feet wide to enclose the largest possible area when placed adjacent to each other, sharing a side.
But hey, don't worry if the math gets too "corralling" – just clown around with the rancher, and everything will work out fine!
To maximize the enclosed area, let's assume that each individual corral has dimensions length (L) and width (W).
Since the two corrals share one side, the length of each corral will be the same (L) and their widths will add up to twice the width of one corral, which is 2W.
The total amount of fencing used for both corrals is 300 feet, so we can set up the equation:
2L + 3W = 300
To find the dimensions that will maximize the enclosed area, we need to express one variable in terms of the other and then substitute it into the area formula.
First, let's express L in terms of W by rearranging the equation:
2L = 300 - 3W
L = (300 - 3W) / 2
Now, let's express the total enclosed area, A, in terms of W:
A = L * W
A = [(300 - 3W) / 2] * W
To maximize the area, we can differentiate the equation with respect to W and set it equal to zero:
dA/dW = 0
Let's compute the derivative and solve the equation:
dA/dW = [(300 - 3W) / 2] - [(3W) / 2]
0 = (300 - 3W) - 3W
0 = 300 - 6W
6W = 300
W = 300 / 6
W = 50
Substituting the value of W into the equation for L, we get:
L = (300 - 3(50)) / 2
L = (300 - 150) / 2
L = 150 / 2
L = 75
Therefore, the dimensions that will yield the largest possible area are each corral having a length (L) of 75 feet and a width (W) of 50 feet.
To find the dimensions that will maximize the enclosed area, we need to use optimization techniques. In this case, we can start by assuming that one corral has a width of "x" and the other corral's width is also "x".
Since each corral is rectangular, we can consider their length to be "y". Since the two corrals are adjacent and share something on one side, their total length is 2y. The fencing required for both corrals will consist of the twice the sum of the lengths and the widths.
The total fencing required is given as 300 feet, so we can write the equation:
2x + 2(2y) = 300
Simplifying the equation, we get:
2x + 4y = 300
Now, we need to express y in terms of x, so we can solve for y and substitute it in the equation for the area of the rectangle.
From the equation 2x + 4y = 300, we get:
4y = 300 - 2x
y = (300 - 2x)/4
y = (150 - x)/2
The total enclosed area (A) is the sum of the areas of both corrals:
A = x * y + x * y
A = 2xy
Now, we can substitute the value of y in terms of x into the equation for the area:
A = 2x * ((150 - x)/2)
A = x * (150 - x)
A = 150x - x^2
To maximize the area, we can take the derivative of A with respect to x and set it equal to zero:
dA/dx = 150 - 2x
Setting 150 - 2x = 0 and solving for x, we find:
150 - 2x = 0
2x = 150
x = 75
Now that we have the value of x, we can substitute it back into the equation for y:
y = (150 - x)/2
y = (150 - 75)/2
y = 75/2
y = 37.5
Therefore, each corral should have a width of 75 feet and a length of 37.5 feet in order to enclose the largest possible area when constructed adjacent to each other.
Rancher whave 3 pieces of fencing of the length L ( two on the border of the fence and one in the middle )
and 2 pieces of fencing of the length W.
3 L + 2 W = 300
Divide both sides by 3
L + 2 W / 3 = 100
Subtract 2 W / 3 to both sides
L = 100 - 2 W / 3
A = L ∙ W = ( 100 - 2 W / 3 ) ∙ W
A = 100 W - 2 W ² / 3
Derivative:
dA / DW = 100 - 2 ∙ 2 W / 3
dA / DW = 100 - 4 W / 3
Function have extreme in point where the derivative is zero.
dA / DW = 100 - 4 W / 3 = 0
Add 4 W / 3 to both sides
100 = 4 W / 3
Multiply both sides by 3
300 = 4 W
300 / 4 = W
75 = W
W = 75 ft
Second derivative test:
If second derivative < 0, then function has a maximum
If second derivative > 0 then function has a minimum
In this case secons derivative = d ( 100 - 4 W / 3 ) / dW = - 4 / 3 < 0
So for W = 75 area has maximum
L = 100 - 2 W / 3
L = 100 - 2 ∙ 75 / 3
L = 100 - 150 / 3
L = 100 - 50
L = 50 ft