optimizatio: 1-km track is to be built with 2 straight sides

[wind]

Hi, can someone help me with this problem

A 1-km track is to be built with 2 straight sides and semicircles at the ends.
a) Find the exact dimensions of the track that encloses the maximum rectangular area. Explain your solution
b) Find the dimensions that enclose the maximum total area. Why do you think tracks are not usually built in this shape?

restriction
0<w<1
0<L<1
(its supposed to be less than or equal to)

A=Lw

P=2L+2piw
1=2L+2piw
(1-2piw)/2=L

A=[(1-2piw)/2]w
A=(w-2piw^2)/2

A'=-4piw/4
A'=-piw
0=w

so o is a possible max or min?

A''=???

help!

Thanks

 

[tkhunny]

Re: optimizatio: 1-km track is to be built with 2 straight s

A=[(1-2piw)/2]w
A=(w-2piw^2)/2

A'=-4piw/4

Better try that derivative again. It should start with "1+" in the parentheses -- and how did it get divided by 4?

Note: Your notation and setup need some work. You will be confusing yourself on this problem without a little more care. Since there are two problems to solve, you must be more clear.

Rectangular Area = L*w

Later, you'll use

Total Area = L*w + pi*(w/2)^2

Keep them separate, distinct and well-defined. Life WILL be better. Trust me on this.

 

[wind]

Keep them separate, distinct and well-defined. Life WILL be better. Trust me on this.

lol, ok

Total Area = L*w + pi*(w/2)^2

sorry but, why is it divided by 2? There are two circles?

so do i keep this

P=2L+2piw
1=2L+2piw
(1-2piw)/2=L

and plug that into this Area = L*w find the value of w then what..?

Thanks

 

[tkhunny]

You were good until you troed to find the derivative. You missed a piece and the 2 magically turned into a 4. Fix those two problems and you're on your way.

 

[wind]

After a 2 week vacation i barly know what I'm doing any more

maximum total area

Total Perimeter

P=2L+ 2pi(w/2)
1=2L+ 2pi(w/2)
1-2L=2pi(w/2)
(1-2L)/2pi=(w/2)
[(1-2L)/2pi]2=w
(1-2L)/pi=w

Total Area

A= Lw • pi(w/2)²
A= L[(1-2L)/pi] • pi([(1-2L)/pi]/2)²
A= (L-2L²)/pi • pi([(1-2L)/pi]1/2)²
A= (L-2L²)/pi • pi((1-2L)/2pi)²
A= (L-2L²)/pi • pi[(1+4L)/2pi]
A= (L-2L²)/pi • (pi+4Lpi)/2pi
A= Lpi +4L²pi -2L²pi -8L³pi/2pi²
A= L +4L² -2L² -8L³/2pi
A= L +2L² -8L³/2pi

A'= [(1 +4L -24L²)(2pi)]/2pi²
A'= (1 +4L -24L²)]/2pi

0= (1 +4L -24L²)]/2pi
0= 1 +4L -24L²

L=-0.137 inadmissable
L≈0.303

(1-0.606)/pi=w
0.1254777≈w

these answers don't seem right...

maximum rectangular area

A=[(1-2piw)/2]w
A=(w-2piw^2)/2

A'=[(1-4piw)(2)-(w-2piw^2)(0)]/2²
A'=[(1-4piw)(2)]/2²
A'=(1-4piw)/2

0=(1-4piw)/2
0=1-4piw
-1=-4piw
-1/-4pi=w

plug w into
(1-2piw)/2=L

 

[TchrWill]

Re: optimizatio: 1-km track is to be built with 2 straight s

A 1-km track is to be built with 2 straight sides and semicircles at the ends.
a) Find the exact dimensions of the track that encloses the maximum rectangular area. Explain your solution
b) Find the dimensions that enclose the maximum total area. Why do you think tracks are not usually built in this shape?

1--W = the track width
2--L = the length of the straight section
3--Pc = the perimeter of the circular ends = PiW
4--L = (3281 - PiW)/2
5--The rectangular area of the track is Ar = W(3281 - PiW)/2 = 3281W - PiW^2)/2
6--dA/dW = (3281 - 2PiW)/2 = 0
7--2PiW = 3281 making W = 522.187m and L = 820.25m.
8--In general terms, L/W = Pi/2 = 1.5708

The maximum enclosed aerea for a given perimeter is a circle.
Circular tracks require a constant leaning of the runner toward the inside of the track thereby using energy and slowing them down.

 

[soroban]

Re: optimizatio: 1-km track is to be built with 2 straight s

Hello, wind!

Let me do part (a) "from scratch".

A 1-km track is to be built with 2 straight sides and semicircles at the ends.

a) Find the exact dimensions of the track that encloses the maximum rectangular area.

                         x
              * * - - - - - - - - * *
          *     |                 |     *
        *       |                 |       *
       *       r|                r|        *
                |                 |
      *         |                 |         *
      *         +                 +         *
      *         |                 |         *
                |                 |
       *       r|                r|        *
        *       |                 |       *
          *     |                 |     *
              * * - - - - - - - - * *
                         x

Let \(\displaystyle x\) = length of the straight track.
Let \(\displaystyle r\) = radius of the semicircular track.

The total length of the track is: \(\displaystyle \:2x\,+\,2\pi r \:=\:1\) km.
. . So we have: \(\displaystyle \:r\:=\:\frac{1\,-\,2x}{2\pi}\;\) [1]

The area of the rectangular field is: \(\displaystyle \:A \:=\x)(2r) \:=\:2xr\;\) [2]

Substitute [1] into [2]: \(\displaystyle \:A \;=\;2x\left(\frac{1\,-\,2x}{2\pi}\right) \:=\:\frac{1}{\pi}\left(x - 2x^2\right)\)

Differentiate, equate to zero, and solve:
. . \(\displaystyle A' \:=\:\frac{1}{\pi}(1\,-\,4x)\:=\:0\;\;\Rightarrow\;\;\fbox{x\,=\,\frac{1}{4}}\)

Substitute into [1]: \(\displaystyle \:r \:=\:\frac{1\,-\,2\left(\frac{1}{4}\right)}{2\pi}\;\;\Rightarrow\;\;\fbox{r\:=\:\frac{1}{4\pi}}\)