Suppose that 430 ft of fencing is used to enclose a corral in the shape of a rectangle with a semicircle whose diameter is a side of the rectangle. (In the figure below, the blue outline represents the fencing.) Find the dimensions of the corral with maximum area.
x=
y=
wkjefbhuisehiufzknse ZJKfznjesfcn eriqv kmwe vojml4 oj n 4wjfjkwe vih
To find the dimensions of the corral with the maximum area, we can use the fact that the perimeter of the corral is 430 ft.
Let's assume the length of the rectangle is x ft, and the width is y ft. Since the diameter of the semicircle is equal to one of the sides of the rectangle, its radius is half the length of the rectangle, so the radius r of the semicircle is (1/2)x ft.
The perimeter of the corral is given by the sum of the lengths of all the sides of the rectangle and half the circumference of the semicircle:
Perimeter = 2x + 2y + (1/2)π(2r) = 430
Simplifying this equation, we get:
2x + 2y + πx = 430
2x + πx + 2y = 430
(2 + π)x + 2y = 430
To find the maximum area of the corral, we need to find the dimensions that maximize the area A. The area A of the corral is given by the sum of the area of the rectangle and half the area of the semicircle:
Area = xy + (1/2)πr^2
Substituting r = (1/2)x, we get:
Area = xy + (1/2)π((1/2)x)^2
Area = xy + (1/8)πx^2
To maximize the area, we need to find the critical points where the derivative of the area with respect to x is equal to zero. Taking the derivative of the area function with respect to x, we get:
dA/dx = y + (1/8)π(2x) = y + (1/4)πx
Setting this derivative equal to zero and solving for y, we get:
y = -(1/4)πx
Substituting this value of y back into the perimeter equation, we get:
(2 + π)x + 2(-(1/4)πx) = 430
(2 + π - 1/2π)x = 430
(4π + 2π^2 - π)x = 860
(4π + 2π^2 - π)x = 860
Simplifying this equation, we get:
(8π + 4π^2 - 2π)x = 1720
(4 + 2π - π)x = 215
(2π + 3)x = 215
x = 215 / (2π + 3)
Substituting this value of x back into the equation for y, we get:
y = -(1/4)π(215 / (2π + 3))
Simplifying this equation, we get:
y = -(215π) / (4(2π + 3))
Hence, the dimensions of the corral with maximum area are:
x = 215 / (2π + 3)
y = -(215π) / (4(2π + 3))
Please note that the subscripts "x" and "y" refer to the dimensions of the corral and do not indicate multiplication.
To find the dimensions of the corral with the maximum area, we can approach it as an optimization problem. The goal is to maximize the area of the corral while using 430 ft of fencing.
Let's break down the problem and find the necessary equations to solve it:
1. Define the variables:
Let x be the width of the rectangle, and y be the length of the rectangle.
2. Determine the equations:
- The perimeter of the rectangle: 2x + y + πy/2 = 430 ft (The sum of the lengths of the four sides of the corral should be equal to 430 ft.)
- The area of the corral: A = xy + πy²/8 (The area of the rectangle plus the area of half of the semicircle)
3. Simplify the equations:
- From the perimeter equation, we can isolate y: y = 430 - 2x - πy/2.
- Substitute the value of y in the area equation: A = xy + πy²/8 = x(430 - 2x - πy/2) + π(430 - 2x - πy/2)²/8.
4. Maximize the area:
To find the maximum area, we will differentiate the area equation with respect to x, set it equal to zero, and solve for x.
- Differentiate the area equation: dA/dx = (430 - 2x - πy/2) + x(-2 - πdy/dx) + π(430 - 2x - πy/2)(-1)(πdy/dx)/(2*8).
- Set dA/dx = 0 and solve for x.
5. Find the corresponding y-value:
Substitute the x-value obtained in step 4 into the perimeter equation and solve for y.
Now, let's solve these equations to find the dimensions of the corral with maximum area.