# Thread: [SOLVED] Optimization Problem argh

1. ## [SOLVED] Optimization Problem argh

Hi everyone!

The Question is:
"A 400km racetrack is to be built with two straight sides and semicircles at the ends. Find the dimensions of the track that encloses the maximum area."
Given:
The two long sides of the rectangle are written with >/= to 100m (each)
The straight side of the 2 semi circles is written with >/= to 20m (each)
(>/= means greater than or equal to, just in case)

Relevant equations are probably:
Area of semicircle = 1/2 * pr2
Area of rectangle = lw

Well i tried but its always the beginning setting up of optimization problems that is the killer, the rest is always easy. It is a study Q for an upcoming test.

2. Originally Posted by falcarius
Hi everyone!

The Question is:
"A 400km racetrack is to be built with two straight sides and semicircles at the ends. Find the dimensions of the track that encloses the maximum area."
Given:
The two long sides of the rectangle are written with >/= to 100m (each)
The straight side of the 2 semi circles is written with >/= to 20m (each)
(>/= means greater than or equal to, just in case)

Relevant equations are probably:
Area of semicircle = 1/2 * pr2
Area of rectangle = lw

Well i tried but its always the beginning setting up of optimization problems that is the killer, the rest is always easy. It is a study Q for an upcoming test.
Falcarius, you are missing ^ with Aof S
$\displaystyle S=pi*(r)^2/2$

and in the end you will have 2 semi circles and 1 rectangle.
Plus you also need to know
$\displaystyle Cs=pi*r$

the total circumprance will be 400km and
$\displaystyle Tc=2*(pi*r)*b$ where b is the base of the rectangle
$\displaystyle Tc=400$km

we want to maximize the total area
$\displaystyle Ta=(pi*r^2)+(2*r*b)$ since the hight of the rectangle is the same as the diamiter or twice the radius of the circle.

with this

$\displaystyle 400/pi=(2*r*b)*pi/pi$
$\displaystyle 400/pi=(2*r*b)$

substitute $\displaystyle 400/pi$ to make
$\displaystyle f(r)=Ta=pi*r^2+400/pi$

from there it's a matter of finding what r is and then finding b

3. oh and a better way of making >/= symbol is to underline > or < like so < or > . I hope this helped

4. Hello, falcarius!

This is a trick question . . . certainly a tricky question.

A 400-km racetrack is to be built with two straight sides and semicircles at the ends.
Find the dimensions of the track that encloses the maximum area.

Given: The two long sides of the rectangle > 100 m.
The diameter of the 2 semicircles >20 m.

I went through the expected procedure for an optimization.
Code:
              * * * * * * * * * * *
*     |               |     *
*       |               |       *
*       r|               |r       *
|               |
*         |               |         *
*         +               +         *
*         |               |         *
|               |
*       r|               |r       *
*       |               |       *
*     |               |     *
* * * * * * * * * * *
L

Let $\displaystyle L$ = length of the straight tracks.
Let $\displaystyle r$ = radius of the semicircular tracks.

The total length of the straight tracks is: $\displaystyle 2L$.
The total length of the curved tracks is: $\displaystyle 2\pi r$
. . Hence, we have: .$\displaystyle 2L + 2\pi r \:=\:400\quad\Rightarrow\quad L \:=\:200 - \pi r$ . [1]

The area of the rectangle is: $\displaystyle 2rL$.
The area of the two semicircles is: $\displaystyle \pi r^2$.
. . Hence, the total area is: .$\displaystyle A \;=\;\pi r^2+ 2rL$ . [2]

Substitute [1] into [2]: .$\displaystyle A \;=\;\pi r^2+ 2r(200 - \pi r)\quad\Rightarrow\quad A \;=\;400r - \pi r^2$

Differentiate and equate to zero: .$\displaystyle A' \;=\;400 - 2\pi r \;=\;0$

. . Hence: .$\displaystyle r \,=\,\frac{200}{\pi}$

Substitute into [1] and we get: .$\displaystyle L \,=\,0$ .
. . . What's going on?

It is fairly well-known that, for a given perimeter,
. . the figure with maximum area is a circle.
So for maximum area, the track should be circular.

But we are told that the straight tracks must be at least 100 meters (each).
. . What do we do now?

It can be shown that, as $\displaystyle L$ increases, the area decreases.
. . Hence, for maximum area, we will use the least $\displaystyle L$, 100.

Under the restrictions, maximum area is attained when $\displaystyle L = 100,\;r = \frac{100}{\pi}$

5. dang I see my mistake. ty man