The figure below depicts a racetrack with ends that are semicircular. The length of the track is 660 ft (1/8 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.)

Well, this is quite the interesting race, isn't it? Okay, I'll give it my best shot.

First, let's break down the problem. We want to find the dimensions of the rectangular portion with the largest possible area. This rectangle is formed by two straight sides and two semicircular ends.

Now, since we have a total track length of 660 ft, we know that the length of each straight side will be half that length, or 330 ft. So the overall length of the rectangle is 330 ft.

Let's focus on the semicircular ends now. Since it's a racetrack, the two ends are identical. So, the sum of their circumferences should be equal to the total perimeter of the rectangle.

The formula for the circumference of a semicircle is C = (π * d) / 2, where d is the diameter. In this case, the diameter will be equal to the length of the rectangle, which is 330 ft.

So, the sum of the circumferences of both semicircular ends is:
C = (π * 330) / 2 = 165π ft

Now, we want to find the radius of each semicircular end, which we'll call r. The formula to find r given the circumference is C = 2πr.

So, 165π = 2πr
r = 165 / 2
r = 82.5 ft

Finally, let's calculate the area of the rectangular portion. The area of a rectangle is given by A = length * width. Since the width is equal to the radius of the semicircular end, our formula becomes:
A = (330 - 2r) * r

Plugging in the value we found for r, we get:
A = (330 - 2 * 82.5) * 82.5
A = 165 * 82.5
A = 13,612.5 ft^2

So, the largest possible area for the rectangular portion of the track is approximately 13,612.5 square feet.

I hope I didn't clown around too much with the calculations!

To find the values of l and r that maximize the area of the rectangular portion of the racetrack, we need to set up an equation and then find the critical points.

Let's start by analyzing the shape of the racetrack. Since the track has ends that are semicircular, we can split it into three parts: two semicircles and a rectangle.

Let the radius of the semicircles be r, and the length of the rectangle be l.

The formula for the area of a rectangle is given by A = length * width.

The length of the rectangle is l, and the width is given by the sum of the diameters of the two semicircles, which is 2r.

Therefore, the formula for the area of the rectangle is A = l * 2r = 2lr.

We also know that the length of the track is 660 ft (1/8 mi). Since the track consists of two semicircles and the rectangle, we can set up the equation:

2r + 2l + πr = 660

Now we need to express l in terms of r and substitute it into the equation for the area, A = 2lr.

From the equation above, we can rearrange it to get:

2l = 660 - 2r - πr
l = (660 - 2r - πr)/2

Now we can substitute this expression for l into the area equation:

A = 2lr = 2 * [(660 - 2r - πr)/2] * r

Simplifying this equation, we have:

A = (660 - 2r - πr) * r

To maximize the area A, we need to find the critical points. We can do that by taking the derivative of A with respect to r and setting it equal to zero.

dA/dr = -2 - π + 4r = 0

Solving this equation for r, we have:

4r = 2 + π
r = (2 + π)/4

Now we can substitute this value of r back into the equation for l:

l = (660 - 2(2 + π)/4 - π(2 + π)/4)/2

l = (660 - (4 + 2π + 2π + π^2)/4)/2

l = (660 - (4 + 4π + π^2)/4)/2

l = (660 - 4 - 4π - π^2)/8

l = (656 - 4π - π^2)/8

Therefore, the values of l and r that maximize the area of the rectangular portion of the racetrack are approximately:

l ≈ (656 - 4π - π^2)/8 ft
r ≈ (2 + π)/4 ft

To find the dimensions that maximize the area of the rectangular portion of the racetrack, we need to use optimization techniques.

Let's assume that the radius of each semicircle is "r" and their combined length is "l". We need to find values for "l" and "r" that maximize the area of the rectangular portion.

First, let's draw the figure to visualize it. The racetrack consists of two straight sides and two semicircular ends:

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The length of the track is given as 660 ft (1/8 mi). We can see that the length of the straight sides is 660 ft - 2r. The length of the semicircular ends is l.

The area of the rectangular portion is given by A = length × width. In this case, the length is the length of the straight sides and the width is the radius of the semicircles.

So, the area of the rectangular portion is A = (660 ft - 2r) × r.

To maximize A, we need to find the critical points where the derivative of A with respect to r is zero. Let's differentiate A with respect to r:

dA/dr = (660 ft - 2r) × 1 - r × 2 = 660 ft - 2r - 2r = 660 ft - 4r.

Setting this derivative equal to zero, we have:

660 ft - 4r = 0.

Solving for r, we find:

4r = 660 ft,
r = 165 ft.

Now that we have the value for r, we can substitute it back into the equation for A to find the corresponding value of l:

A = (660 ft - 2r) × r = (660 ft - 2(165 ft)) × 165 ft = 330 ft × 165 ft = 54,450 ft^2.

Thus, the maximum area of the rectangular portion of the racetrack is 54,450 ft^2, when the radius of each semicircle is 165 ft.

2l + 2πr = 660

a = 2lr + πr^2
= 2(660-2πr)/2 * r + πr^2
= r(660-2πr) + πr^2
= 660r - πr^2

Now just set da/dr=0 to find r, then l.