A rancher wants to construct two identical rectangular corrals using 300 ft of fencing. The rancher decides to build them adjacent to each other, so they share fencing on one side. What dimensions should the rancher use to construct each corral so that together, they will enclose the largest possible area?
easy way. Let the length be 2x and the width be y.
Then 2x=3y=150 maximizes the area
The corral will be 150 by 50 (two 75x50 corrals)
Or, do the math
2x+3y = 300
y = 100 - 2x/3
area a = 2xy = 2x(100 - 2x/3) = 200x - 4/3 x^2
Max area at the vertex of that parabola. Use algebra or calculus.
Well, to maximize the area, the rancher needs to think outside the box – or in this case, outside the corral! By forming the two rectangular corrals adjacent to each other, they'll share one side. This means that the total length of shared fencing will be deducted twice from the total perimeter.
Let's call the length of the shared side 'x' and the length of the side perpendicular to it 'y'. Since there are two identical corrals, each corral's perimeter will be as follows:
2x + y + y = 300
Simplifying the equation, we get:
2x + 2y = 300
2(x + y) = 300
x + y = 150
Now, let's focus on finding the area. The area of each corral will be:
Area = x * y
We need to find the values of 'x' and 'y' that will maximize this area. Since we already know that x + y = 150, we can rewrite it to solve for x:
x = 150 - y
Substituting this into the area equation, we get:
Area = (150 - y) * y
To maximize the area, we need to find the value of y that gives us the maximum result. So, let's put on our clown hats and solve this! *honk honk*
Calculating the derivative of the area equation with respect to y, we get:
d(Area)/dy = 150 - 2y
Setting this derivative equal to zero, we can find the value of y that maximizes the area:
150 - 2y = 0
2y = 150
y = 75
Plugging this value back into the equation x + y = 150, we find:
x + 75 = 150
x = 75
So, to enclose the largest possible area, each corral should have dimensions of 75 ft by 75 ft. You can call them the "Jumbo Double-Sized Corrals" – perfect for clown cows and acrobatic horses! *insert circus music*
To find the dimensions that will enclose the largest possible area, we can use the derivative of the area formula with respect to one of the variables. Let's call the length of each corral L and the width W.
Since the two corrals share fencing on one side, the total length of shared fencing will be the sum of the lengths of the two adjacent sides of the two rectangular corrals. Therefore, the length of the shared fencing is: L + L = 2L.
The remaining fencing will be used for the other three sides of each corral. So, the total length of the other three sides is: W + W + L + L = 2W + 2L.
Given that there are 300 ft of fencing available, we can equate the total length of the other three sides to 300 ft and solve for W:
2W + 2L = 300.
Rearranging the equation, we have:
W + L = 150 - Equation 1.
The area of each corral is given by the formula: A = L * W.
Now, we need to express the area in terms of one variable so that we can determine the maximum value. Using Equation 1, we can express L in terms of W:
L = 150 - W.
Substituting this value of L into the area formula, we have:
A = W * (150 - W).
Expanding the equation, we get:
A = 150W - W^2.
To find the maximum value of A, we can take the derivative of A with respect to W and set it equal to zero:
dA/dW = 150 - 2W = 0.
Solving this equation for W, we find:
2W = 150,
W = 75.
Substituting this value of W back into Equation 1, we can solve for L:
L = 150 - W = 150 - 75 = 75.
Therefore, each corral should have dimensions of 75 ft by 75 ft to enclose the largest possible area when constructed adjacent to each other.
To find the dimensions that will enclose the largest possible area, we can use the concept of optimization.
Let's start by representing the dimensions of one rectangular corral. Let's assume the length of the corral is L and the width is W. Since the two corrals will share fencing on one side, we can form one large rectangle with length 2L and width W.
Now, let's calculate the perimeter of this large rectangle by adding up the lengths of all the sides:
Perimeter = 2L + L + 2W + W = 3L + 3W
We can also use the information in the problem statement, which states that the rancher has 300 ft of fencing:
Perimeter = 300
Substituting the given information:
3L + 3W = 300
Next, we need to represent the area of the large rectangle. The area of a rectangle is calculated by multiplying its length and width:
Area = (2L)(W) = 2LW
To find the dimensions that maximize the area, we need to maximize the value of 2LW. To do that, we can solve the equation we previously derived for one side's perimeter (3L + 3W = 300) and represent one variable in terms of the other:
3L + 3W = 300
L + W = 100 (Dividing the equation by 3)
Let's represent L in terms of W:
L = 100 - W
Now, substitute L in the formula for the area:
Area = 2(100 - W)W = 200W - 2W^2
To find the maximum area, we can find the maximum point of this quadratic function. We can do this by finding the vertex of the quadratic equation.
The vertex of a quadratic function in the form f(x) = ax^2 + bx + c is given by the formula:
x = -b / 2a
In this case, a = -2 and b = 200. Substituting these values in the formula, we can find the value of W that maximizes the area:
W = -200 / (2*(-2)) = -200 / (-4) = 50
Therefore, the width of each corral is 50 ft.
Substituting the value of W back into the equation L + W = 100, we can find the length of each corral:
L + 50 = 100
L = 100 - 50
L = 50
Therefore, the length of each corral is also 50 ft.
So, the rancher should construct each corral with dimensions 50 ft by 50 ft to enclose the largest possible area.