# 2 math question help

• December 5th 2008, 03:26 AM
juzo
2 math question help
1)
A balloon is rising vertically upward at 22m/s. If jane is standing 45 metres away from where the balloon is lifted, how fast is the balloon flying away from jane at the moment when the balloon is 70metres high from the field?

2)
A cylindrical container , with the top ended is opened, is to be constructed to hold a volume of 1.50m³ . Find the dimensions of the container such that the materials used would be a minimum. Give your answers to 2 decimal places.
• December 5th 2008, 04:23 AM
earboth
Quote:

Originally Posted by juzo
...

2)
A cylindrical container , with the top ended is opened, is to be constructed to hold a volume of 1.50m³ . Find the dimensions of the container such that the materials used would be a minimum. Give your answers to 2 decimal places.

Let r denote the radius of the base and h the height of the cylinder. The you know:

$V = \pi\cdot r^2 \cdot h$

and the area of the material which is used to build the cylinder:

$a = \pi r^2 + 2\pi\cdot r \cdot h$

Solve the equation of the volume for h and plug in this term into the equation of a:

$h = \dfrac{V}{\pi r^2}$

$a(r)=\pi r^2 + 2\pi\cdot r \cdot \dfrac{V}{\pi r^2} = \pi r^2 + \dfrac{2V}{r}$

Determine the first derivation and solve a'(r) = 0 for r:

$2\pi r - \dfrac{2V}{r^2} = 0 ~\implies~r=\sqrt[3]{\dfrac V\pi}$
• December 5th 2008, 07:42 PM
juzo
Ooo. Thx. Does any1 noe abt question 1 too???
• December 5th 2008, 08:43 PM
Chop Suey
For question number one, let us first list the given. I will assume that y and x corresponds for the vertical and horizontal direction, respectively. I will also assume that Jane is a stationary object ("Jane is standing...").

$\frac{dy}{dt} = 22~m/s$

$y = 70~m$

$\frac{dx}{dt} = 0$

$x = 45~m$

$\frac{dz}{dt} = ?$

$z = \sqrt{x^2+y^2} = \ldots$

Note that the distance between the balloon and Jane is the hypotenuse z of the right-triangle with legs x and y. We are trying to find $\frac{dz}{dt}$.

We've got everything in the bag. Differentiating the pythagorus:
$x^2 + y^2 = z^2$

$2x\frac{dx}{dt} + 2y\frac{dy}{dt}=2z\frac{dz}{dt} \implies \frac{dz}{dt} = \frac{\displaystyle x\frac{dx}{dt} + y\frac{dy}{dt}}{z}$

And we're done. I have showed you how to solve it, you do the computations. ;)