# related rates and optimization problems

• Apr 21st 2008, 10:56 AM
qu2010
related rates and optimization problems
1.A spotlight is on the ground 100 feet from a building that has vertical sides. A person 6 feet tall starts at the spotlight and walkd directly towards the building at a rate of 5 feet per second.
a) how fast is the top of the persons shadow moving down the building when the person is 50 feet away.
b) how fast is the top of the persons shadow moving down the building when the person is 25 feet away.

2. A spherical ball 8 inches in diameter is coated with a layer of ice of uniform thickness. If the ice melts at the rate of 10in3/min, how fast is the thickness of the ice descresing whe it is 2 inches thick?

3. An open box with a capcity of 36,000 cubic inches is to be twice as long as it is wide. the material for the box costs 10 cents per square foot. what are the dimensions of the least expensive box? how much does it cost?
• Apr 21st 2008, 12:28 PM
Soroban
Hello, qu2010!

Quote:

1. A spotlight is on the ground 100 feet from a vertical building.
A 6-foot man starts at the spotlight and walks towards the building at 5 ft/sec.

Code:

C *
|  *  A
|      *
y |      |  *
|      |6      *
|      |          *
* - - - * - - - - - - - * - -
D  x  B    100-x    L
: - - - - 100 - - - - - :

The spotlight is at $\displaystyle L.$
The man is: .$\displaystyle AB = 6$
His shadow is: .$\displaystyle y = CD$
His distance from the building is: .$\displaystyle x \,=\,DB$
. . Hence: .$\displaystyle BL \:=\:100-x$
We are given: .$\displaystyle \frac{dx}{dt} = -5$ ft/sec.

From similar right triangles $\displaystyle CDL\text{ and }ABL\text{ we have: }\:\frac{y}{100} \:=\:\frac{6}{100-x}$

. . Then: .$\displaystyle y \:=\:600(100-x)^{-1}$

Differentiate with respect to time: .$\displaystyle \frac{dy}{dt} \:=\:-600(100-x)^{-2}\left(-\frac{dx}{dt}\right)$

. . and we have: .$\displaystyle \boxed{\frac{dy}{dt} \;=\;\frac{600}{(100-x)^2}\,\frac{dx}{dt}}$

Quote:

a) How fast is the top of the man's shadow moving down the building
when he is 50 feet from the building?

When $\displaystyle x = 50$ . . .

. . $\displaystyle \frac{dy}{dt} \;=\;\frac{600}{(100-50)^2}\,(-5) \;=\;\boxed{-\frac{6}{5}\text{ ft/sec}}$

Quote:

b) How fast is the top of the man's shadow moving down the building
when he is 25 feet from the building?

When $\displaystyle x = 25$ . . .

. . $\displaystyle \frac{dy}{dt} \;=\;\frac{600}{(100-25)^2}\,(-5) \;=\;\boxed{-\frac{8}{15}\text{ ft/sec}}$