# Optimizing the area of a rectangle inside a racetrack

1. Apr 15, 2015

### dlp248

1. The problem statement, all variables and given/known data

Maximize the area (in feet) of the rectangular field inside of a mile long racetrack.

2. Relevant equations

Circumference of a circle = 2πr
P= 2x + 2y

3. The attempt at a solution

Area of the semicircles = πr^2
Area of the rectangle = 2rh

A(r) = πr^2 +2rh
P= 2πr + 2h + 4r

5280ft = 2πr + 2h + 4r

h= 2640 - πr - 2r

A(r) = πr^2 + 2r(2640 - πr - 2r)
A(r) = πr^2 + 5280r - 2πr^2 - 4r^2
A(r) = 5280r - πr^2 - 4r^2

A'(r) = 5280 - 2πr -8r
0 = 5280 - 2πr -8r
2πr + 8r = 5280
r(2π + 8) = 5280

r = 5280/(2π + 8) ft
r ≈ 369.67 ft

A"(r) = -2π - 8
A"(r) < 0
r is a maximum

h = 2640 - (5280/(2π + 8))π - 2(5280/(2π + 8))
h = 10560/(2π + 8) ft
h ≈ 739.33 ft

A(r) = πr^2 +2rh
A(r) = π(5280/(2π + 8) ft)^2 + 2((5280/(2π + 8) ft)(10560/(2π + 8) ft))
A(r) ≈ 975,917 ft^2

I believe this answer is incorrect because for the perimeter, I included the two diameters of the semicircles (the width of the imaginary rectangle). So I ended up using part of the given distance to make these sides and my perimeter formula does not represent the actual perimeter of the track. I tried to redo this problem just using:

P= 2πr + 2h

however this formula produces an r value that makes the h value equal zero. Obviously I am missing a small detail and I have no idea what it is!

Last edited: Apr 15, 2015
2. Apr 15, 2015

### SteamKing

Staff Emeritus
This is one of those problems where a simple sketch can go a long way to clarify what needs to be done to find a solution.

Note: you can also express the circumference of a circle using the diameter, which is also one of the dimensions of the rectangle. It's not clear why you included the area of the semicircles in your calculations, since the OP explicitly specified the area of the rectangular field in side the track is the quantity to be maximized.

You went to great pains to show your calculations for an obviously incorrect solution, but you have omitted the calculations which supposedly use the correct formula for the perimeter of the race track. We can't help unless you show your latest work.

3. Apr 15, 2015

### dlp248

I am not sure how to add a sketch to this post; It is a rectangle with two semicircle ends. This is the rest of my work:

P= 2πr + 2h

5280 = 2πr + 2h
h = 2640 - πr

A(r) = πr^2 + 2rh

A(r) = πr^2 + 2r(2640 - πr)
A(r) = πr^2 + 5280r - 2πr^2
A(r) = 5280r - πr^2

A'(r) = 5280 - 2πr
0 = 5280 - 2πr
r = 5280/2π
r = 2640/π

h = 2640 - π(2640/π)
h = 0

Now that you mention that I am using the area of the circle in my formula I think I see where I am going wrong. I am definitely over thinking this. I wonder if I should be using these two formulas instead of the others I have been using:

P= 2πr + 2h
A(r) = 2rh

Using
h = 2640 - πr

I just need to plug this h value into the area of the rectangle, differentiate, and solve for r. Let me know if that is on the right track. Thank you!

4. Apr 15, 2015

### dlp248

Here is my attempt at a sketch:

#### Attached Files:

• ###### 20150415_181057.jpg
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5. Apr 15, 2015

### SteamKing

Staff Emeritus
Try it and see what you get.