A horse breeder wants to construct a corral next to a horse barn that is L=16 feet long, using the barn as part of one side of the corral as shown in the figure above. The breeder has 280 feet of fencing available.
Find the value of x which maximizes the amount of area the corral can enclose
no figure?
Since 16 feet of fencing is not used, it is the same as if a complete rectangle of perimeter 296 were being built.
The maximum area will be for a square of side 296/4 = 74 feet.
To work it out, let x be the side with the barn in it. Then we have
2x-16+2y=280
x+y=148
the area is
a = xy = x(148-x) = 148x-x^2
the vertex is at x = -b/2a = 74
To find the value of x that maximizes the area the corral can enclose, we need to maximize the area function with respect to x.
Let's start by understanding the problem and drawing a diagram. The corral is in the shape of a rectangle, with one side formed by the horse barn. Let's assume the width of the corral is y feet and the length of the corral parallel to the barn is x feet.
From the problem, we know that the barn is 16 feet long and the total length of fence available is 280 feet. We are trying to find the value of x, which maximizes the area of the corral.
Let's break down the problem into steps:
Step 1: Define the variables:
- Length of the barn: L = 16 feet
- Length of the corral parallel to the barn: x feet
- Width of the corral: y feet
- Total length of fencing available: P = 280 feet
Step 2: Determine the perimeter equation:
Since the corral consists of two sides of length x and two sides of length y, the perimeter of the corral is given by:
P = x + y + y = x + 2y
We are given that P = 280, so we can write the perimeter equation as:
280 = x + 2y
Step 3: Express the area of the corral in terms of x and y:
The area of a rectangle is given by the product of its length and width:
A = x * y
Step 4: Solve the perimeter equation for y in terms of x:
To express the width y in terms of x, we rearrange the perimeter equation:
2y = 280 - x
y = (280 - x) / 2
Step 5: Substitute the value of y into the area equation:
Substituting the expression for y into the area equation, we get:
A = x * ((280 - x) / 2)
Step 6: Simplify the area equation:
To simplify the area equation, we multiply x by both terms in the bracket:
A = (280x - x^2) / 2
Step 7: Find the derivative of the area equation:
To maximize the area, we need to find the critical points of the area function. We can do this by finding the derivative of the area equation with respect to x and setting it equal to zero:
dA/dx = (280 - 2x) / 2
(280 - 2x) / 2 = 0
Step 8: Solve for x:
Setting the derivative equal to zero and solving for x, we get:
280 - 2x = 0
2x = 280
x = 140
Step 9: Find the corresponding y:
Now that we have the value of x, we can substitute it back into the perimeter equation to find the value of y:
280 = 140 + 2y
2y = 280 - 140
2y = 140
y = 70
Step 10: Calculate the maximum area:
Finally, we substitute the values of x and y into the area equation to find the maximum area:
A = (280 * 140 - 140^2) / 2
A = 39200 / 2
A = 19600 square feet
Therefore, the value of x that maximizes the area the corral can enclose is 140 feet, and the maximum area is 19600 square feet.