# Is there anyone alive who can work this out ?

• June 13th 2006, 01:32 PM
mayster17
Is there anyone alive who can work this out ?
A solar station is to be built at ground level on the east-west line between two buildings. The distance between the two buildings is 100 meters, the height of the taller building is 80 meters, and the height of the smaller building is 40 meters. How far from the taller building should the station be placed in order to maximize the
number it will be in the sun on a day when the sun passes directly overhead? Note that :

(theta) = (pi) - (cot)^(-1) [(100-x)/(4)]

and find the value of x which maximizes theta.
• June 13th 2006, 02:52 PM
Soroban
Hello, mayster17!

I'll give it a try . . .

Quote:

A solar station is to be built at ground level on the east-west line between two buildings.
The distance between the two buildings is 100 meters.
The height of the taller building is 80 m; the height of the smaller building is 40 m.
How far from the taller building should the station be placed in order to maximize
the time it will be in the sun on a day when the sun passes directly overhead?

I don't agree with that equation . . .
Code:

      *       |\       | \       |  \       |  \              *       |    \            /|       |    \          / |   80 |      \        /  |       |      \      /  | 40       |        \    /    |       |        \ θ /    |       |        α \ / β    |     - * - - - - - * - - - * -             x      100-x
We have: . $\theta\;=\;\pi - \alpha - \beta \;= \;\pi - \cot^{-1}\left(\frac{x}{80}\right) - \cot^{-1}\left(\frac{100-x}{40}\right)$

Then: . $\frac{d\theta}{dx}\;=\;\frac{\frac{1}{80}}{1 + \left(\frac{x}{80}\right)^2} + \frac{-\frac{1}{40}}{1 + \left(\frac{100-x}{40}\right)^2} \;= \;0$

Simplify: . $\frac{80}{x^2 + 80^2} - \frac{40}{(100-x)^2 + 40^2}\;=\;0$

. . . $80[((x-100)^2 + 40^2] \;= \;40[x^2 + 80^2]\quad\Rightarrow$ . $2[(x-100)^2+40^2] \;=$ $\;x^2+80^2$

Simplify: . $x^2 - 400x + 16,800\;=\;0$

Quadratic Formula: . $x\;=\;\frac{-(-400) \pm\sqrt{(-400)^2 - 4(16,800)}}{2} \;= \;\frac{400\pm \sqrt{92,800}}{2}
$

Therefore: . $x\;=\;200 - 20\sqrt{59}\;\approx\;46.4\text{ meters.}$