A rancher is fencing off a rectangular area with a fixed perimeter of 76 m. What dimensions would yield the maximum area? Find the maximum area.
To find the dimensions that yield the maximum area, we can use the concept of calculus. Let's denote the length of the rectangle as L and the width as W.
Given that the perimeter is fixed at 76 m, we can write the equation:
2L + 2W = 76
Simplifying the equation, we have:
L + W = 38
Now, let's express the area, A, in terms of L and W:
A = L * W
To find the maximum area, we need to find the values of L and W that satisfy the perimeter equation while maximizing the area.
To do this, we can solve the perimeter equation for one variable and substitute into the area equation. Let's solve the perimeter equation for W:
W = 38 - L
Substituting this into the area equation, we get:
A = L * (38 - L)
Expanding the equation, we have:
A = 38L - L^2
To find the maximum area, let's differentiate the area equation with respect to L:
dA/dL = 38 - 2L
To find the maximum, we set the derivative equal to zero and solve for L:
38 - 2L = 0
2L = 38
L = 19
Now, substitute L = 19 back into the perimeter equation to find W:
W = 38 - L
W = 38 - 19
W = 19
Therefore, the dimensions that yield the maximum area are L = 19m and W = 19m.
To find the maximum area, substitute the values of L and W into the area equation:
A = L * W
A = 19 * 19
A = 361 square meters
So, the maximum area is 361 square meters.
To find the dimensions that would yield the maximum area, we need to determine the length and width of the rectangular area. Let's assume the length is "L" and the width is "W".
We are given that the perimeter is fixed at 76 meters. Perimeter of a rectangle is given by the formula P = 2L + 2W, where L is the length and W is the width. So, we can write the equation as:
76 = 2L + 2W
Now, we want to find the dimensions that yield the maximum area. The area of a rectangle is given by the formula A = L * W.
To find the maximum area, we can express one variable in terms of the other variable. In this case, we can express W in terms of L using the equation for the perimeter:
2L + 2W = 76
2W = 76 - 2L
W = (76 - 2L)/2
W = 38 - L
Now, substitute this value of W into the equation for the area:
A = L * W
A = L * (38 - L)
A = 38L - L^2
To find the maximum area, we need to find the value of L that maximizes the area. We can do this by finding the vertex of the parabola formed by the equation for the area.
The vertex of a parabola represented by the equation y = ax^2 + bx + c is given by the x-coordinate:
x = -b / (2a)
In our equation, a = -1, b = 38, and c = 0. So, plug these values into the formula to get the x-coordinate:
L = -38 / (2*(-1))
L = -38 / -2
L = 19
Now, substitute this value of L back into the equation for W:
W = 38 - L
W = 38 - 19
W = 19
So, the dimensions that yield the maximum area are L = 19 meters and W = 19 meters.
To find the maximum area, substitute these values of L and W into the equation for the area:
A = L * W
A = 19 * 19
A = 361
Therefore, the maximum area of the fenced rectangular area is 361 square meters.