The figure below depicts a racetrack with ends that are semicircular. The length of the track is 880 ft (1/6 mi). Find l and r so that the area of the rectangular portion of the region enclosed by the racetrack is as large as possible. (Round your answers to the nearest foot.)
To find the values of l and r that maximize the area of the rectangular portion of the racetrack, we need to use optimization techniques.
We can start by drawing a diagram of the racetrack, which consists of two semicircles and a rectangle:
```
Semi-circle 1 Semi-circle 2
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_______________________
```
Let's denote the radius of each semicircle as r, and the length of the rectangle as l.
The total length of the racetrack is given as 880 ft, which is equal to the circumference of the two semicircles plus the length of the rectangle:
Circumference of semi-circle 1 + Circumference of semi-circle 2 + Length of rectangle = 880 ft
The formula for the circumference of a circle is given by 2πr, but since we're dealing with semicircles, we only need half of that:
πr + πr + l = 880 ft
2πr + l = 880 ft
Now, we need to express l in terms of r, so we can rewrite the equation as:
l = 880 ft - 2πr
The area of the rectangular portion of the racetrack can be calculated by multiplying the length and width of the rectangle, which is l and r respectively:
Area = l * r
Area = (880 ft - 2πr) * r
To find the maximum area, we can take the derivative of the area function with respect to r, set it equal to zero, and solve for r.
d(Area)/dr = (880 ft - 2πr) * 1 + r * (-2π) = 0
Simplifying this equation, we get:
880 ft - 2πr - 2πr = 0
880 ft - 4πr = 0
Solving for r, we have:
4πr = 880 ft
r = 880 ft / 4π
r ≈ 70.23 ft
Now, substitute this value of r back into the equation for l:
l = 880 ft - 2πr
l = 880 ft - 2π * 70.23 ft
l ≈ 566.69 ft
Therefore, to maximize the area of the rectangular portion of the racetrack, the values of l and r should be approximately 567 ft and 70 ft respectively.
To find the values of l and r that maximize the area of the rectangular portion of the racetrack, we need to use calculus to find the maximum.
Let's start by drawing the figure described in the problem:
```
l
+---------------------------+
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| Racetrack |
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+ Semicircular End +
radius r
```
From the problem, we know that the length of the track is 880 ft (1/6 mi), which means the length of the rectangular portion (l) plus the curved portion (the two semicircular ends) must equal 880 ft.
We are also told that the curved ends of the racetrack are semicircular, so their combined length will be equal to the circumference of a circle with radius r, which is 2πr.
So, the equation for the length of the track is:
l + 2πr = 880
Next, the area of the rectangular portion of the racetrack is given by:
Area = length * width
We know that the width of the rectangular portion is twice the radius of the semicircular end, so:
Width = 2r
Therefore, the area of the rectangular portion is:
Area = l * 2r
To simplify the problem, let's solve the equation for l:
l = 880 - 2πr
Now, substitute this value of l into the area equation:
Area = (880 - 2πr) * 2r
To find the maximum area, we need to differentiate the area equation with respect to r, set it equal to zero, and solve for r:
d(Area)/dr = 0
Differentiating the equation:
d(Area)/dr = 880 - 4πr = 0
Solve for r:
4πr = 880
r = 220/π
Now, substitute this value of r back into the equation for l:
l = 880 - 2π(220/π) = 880 - 440 = 440
So, the values that maximize the area of the rectangular portion of the racetrack are l = 440 ft and r = 220/π ft (approximately 70.08 ft).
Therefore, the rectangular portion of the racetrack has dimensions 440 ft by 2(220/π) ft, which is approximately 440 ft by 140.17 ft.
Please note that the values have been rounded to the nearest foot as per the instructions.