# application of differentiation

• October 29th 2009, 08:17 AM
gilliana
[SOLVED]application of differentiation
hello everyone, i have this question which i am unsure what to start with..

A pole a+b meters high stands vertically in a river b meters deep. Assume a > b. How many meters above the surface of the water must it break so that the top bending down touches the bottom of the river and the horizontal distance on the surface of the water between the points where the parts of the pole enter the water would be a maximum?

Suppose that the break point is x meters above the surface of water. Specify the function that is to be maximized in terms of x.

all i can see is the height is x + b..
any help would be appreciated.
• October 29th 2009, 11:07 AM
Soroban
Hello, gilliana!

I came up with a very messy function.

Quote:

A pole $a+b$ m high stands vertically in a river $b$ meters deep.
Assume a > b.

How many meters above the surface of the water must it break
so that the top bending down touches the bottom of the river
and the horizontal distance on the surface of the water between the points
where the parts of the pole enter the water would be a maximum?

Suppose that the break point is $x$ meters above the surface of water.
Specify the function that is to be maximized in terms of x.

Code:

      *       |       |       |       |     a |       |                P *       |                  |\       |                  | \       |                x |  \       |                  |  \       |                  |  θ \    a-x       * ~ ~ ~          Q * - - * S       |                  |  y  \       |                  |      \     b |                b |        \       |                  |        \       |                  | v      θ \       * - - - -        R * - - - - - * T                                 z

The pole is broken at $P\!:\;\;PQ = x,\;PT = a-x,\;QR = b$
Let $y = QS,\;z = RT$
Let $\theta = \angle PSQ = \angle PTR$

In right triangle $PRT\!:\;\tan\theta = \frac{x+b}{z}$
In right triangle $PQS\!:\;\tan\theta =\frac{x}{y}$
. . Hence: . $\frac{x+b}{z} \:=\:\frac{x}{y} \quad\Rightarrow\quad y \:=\:\frac{xz}{x+b}$ .[1]

In right triangle $PRT\!:\;\;z^2 + (x+b)^2 \:=\:(a-x)^2$

. . Solve for $z\!:\;\;z \:=\:\sqrt{a^2-b^2 - 2(a+b)x}$

$\text{Substitute into }{\color{blue}[1]}\!:\;\;y \;=\;\frac{x\left[a^2-b^2-2(a+b)x\right]^{\frac{1}{2}}}{x+b}$

And that is the function to be maximized.

• October 29th 2009, 04:05 PM
gilliana
thank you very much Soroban, it is a nice clear diagram that explains everything.