Suppose that 404 ft of fencing are used to enclose a corral in the shape of a rectangle with a semicircle whose diameter is a side of the rectangle.

Find the dimensions of the corral with maximum area.
x=........ft
Y=......ft

rectangle perimter = x + pi*x + 2h

h = (404 - x - pi*x)/2

area = xh + 1/2 pi*(x/2)^2
= x(404 - x - pi*x)/2 + 1/8 pi*x^2
= 202x - x^2/2 - pi*x^2/2 + pi*x^2/8
= -(3pi/8 + 1/2)x^2 + 202x
= x(202 - (3pi/8 + 1/2)x)

This has a maximum when x = 60.187

You can probably get the dimensions from this.

To find the dimensions of the corral with maximum area, we need to optimize the area function. First, let's define the variables:

Let's call the length of the rectangle x and the width of the rectangle y. Since the semicircle's diameter is equal to one side of the rectangle, the width of the rectangle (y) will also be the radius of the semicircle.

Now, let's write the equation for the perimeter of the corral:

Perimeter = 2x + (πy)/2 + 2y = 404 ft

Simplifying the equation, we have:

2x + πy + 4y = 404
2x + 5y = 404 (Equation 1)

Next, let's write the equation for the area of the corral:

Area = x*y + (πy^2)/2

We want to maximize the area, which means finding the maximum value of the area function. To do this, we need to find a way to express the area in terms of a single variable. We can use Equation 1 to express x in terms of y:

2x = 404 - 5y
x = (404 - 5y)/2

Substituting this value of x into the area equation, we get:

Area = ((404 - 5y)/2) * y + (πy^2)/2
Area = (404y - 5y^2)/2 + (πy^2)/2
Area = (404y + πy^2 - 5y^2)/2

To find the maximum area, we need to find the critical points of the area function. We can do this by finding the derivative of the area function with respect to y, setting it equal to zero, and solving for y.

Let's differentiate the area function:

d(Area)/dy = (404 + 2πy - 10y)/2

Setting it equal to zero and solving for y:

(404 + 2πy - 10y)/2 = 0
404 + 2πy - 10y = 0
2πy - 10y = -404
2y(π - 5) = -404
y(π - 5) = -202
y = -202/(π - 5)

Since the width of the corral cannot be negative, we can disregard the negative solution. Therefore, we have:

y = -202/(π - 5)

To find the value of x, we can substitute this value of y back into Equation 1:

2x + 5(-202/(π - 5)) = 404
2x - 1010/(π - 5) = 404
2x = 404 + 1010/(π - 5)
x = (404 + 1010/(π - 5))/2

At this point, we can calculate the values of x and y using the above equations.

To find the dimensions of the corral with maximum area, we can set up an equation using the given information.

Let's assume the length of the rectangle is 'x' feet, and the width of the rectangle is 'y' feet. We also know that the semicircle has a diameter equal to the width of the rectangle, which makes its radius equal to half of the width.

Now, let's calculate the perimeter of the corral:

Perimeter = Length + Width + Circumference of the semicircle

Given that the perimeter is 404 ft, we can write the equation:

404 = x + y + πr

Since the radius equals half of the width, we can substitute 2r with y:

404 = x + y + πy

Next, let's find the area of the corral:

Area = Area of the rectangle + Area of the semicircle

The area of the rectangle is x * y, and the area of the semicircle is 0.5 * πr^2, where r equals half of the width:

Area = xy + 0.5 * π(r^2)

Substituting r with y/2, we get:

Area = xy + 0.5 * π(y/2)^2
= xy + 0.5 * π(y^2/4)
= xy + 0.125 * πy^2

Now, we need to maximize the area. To do this, we can use calculus and take the derivative of the area with respect to y:

dA/dy = x + 0.25 * πy

To find the maximum area, we can set the derivative equal to zero:

x + 0.25 * πy = 0

Solving for x, we get:

x = -0.25 * πy

Substituting this value of x in the perimeter equation, we have:

404 = -0.25 * πy + y + πy
404 = 0.75 * πy + y
404 = (0.75π + 1)y

Now, we can solve for y:

y = 404 / (0.75π + 1)

Once we have the value of y, we can substitute it back into the equation for x:

x = -0.25 * πy

Finally, we can calculate the dimensions of the corral:

x = -0.25 * π * (404 / (0.75π + 1))
y = 404 / (0.75π + 1)

Simplifying the values of x and y:

x ≈ -126.79 ft
y ≈ 317.97 ft

Therefore, the dimensions of the corral with maximum area are approximately:

x ≈ -126.79 ft
y ≈ 317.97 ft