# 2 Optimization Problems

• Nov 4th 2007, 04:49 PM
Super Mallow
2 Optimization Problems
Looking for some urgent help here on these 2 optimization problems

1) A right triangle whose hypotenuse is SqureRoot3 m long is revolved about one of its legs to generate a right circular cone. Find the radius, height, and volume of the cone of the greatest volume that can be made this way

2) The 8 foot wall shown here stands 27 feet from the building. Find the length of the shortest straight beam that will reach the side of the building from the ground outside the wall

http://img100.imageshack.us/img100/2...problemyz4.jpg
• Nov 4th 2007, 06:30 PM
Soroban
Hello, Super Mallow!

Here's some help with #1. . .

Quote:

1) A right triangle whose hypotenuse is $\displaystyle \sqrt{3}$ m long is revolved about one
of its legs to generate a right circular cone. Find the radius, height, and
volume of the cone of the greatest volume that can be made this way

Code:

. . . - - - - - - * . . . - - - - - * | . . . - - √3  *  | . . . - - - *    |√3sinθ . . . - - *      | . . . - * θ      |       * - - - - - *         √3cosθ
This triangle will be revolved about its vertical side.

The volume of a cone is: .$\displaystyle V \;=\;\frac{\pi}{3}r^2h$

Let $\displaystyle \theta$ be the left acute angle.
Then: .$\displaystyle r \,=\,\sqrt{3}\cos\theta,\;h\,=\,\sqrt{3}\sin\theta$

We have: .$\displaystyle V \;=\;\frac{\pi}{3}\left(\sqrt{3}\cos\theta\right)^ 2\left(\sqrt{3}\sin\theta\right) \;=\;\sqrt{3}\pi\sin\theta\cos^2\!\theta$

Can you finish it now?

• Nov 5th 2007, 06:10 AM
Super Mallow
..I don't understand how you got that at all actually with the equation...

With that formula, do I take hte derivative and find the criticals?
• Nov 5th 2007, 09:02 AM
ticbol
) The 8 foot wall shown here stands 27 feet from the building. Find the length of the shortest straight beam that will reach the side of the building from the ground outside the wall

In the posted figure,
Let H = height of the top end of the beam
And x = distance of the bottom end of the beam from the foot of the 8-ft wall.

By proportion,
8/x = H/(x+27)
H = 8(x+27)/x

Let B = length of the straight beam.
By Pythagorean theorem,
B^2 = (x+27)^2 +H^2
B^2 = (x+27)^2 +[8(x+27)/x]^2
B^2 = [(x+27)^2]*[(1 +64/(x^2)]
B^2 = [(x+27)^2]*[(x^2 +64)/x^2]
Take the sqrt pof both sides,
B = [(x+27)/x]*sqrt(x^2 +64) --------------(i)

Differentiate both sides of (i) with respect to x,
dB/dx = [(x+27)/x]*[x /sqrt(x^2 +64)] +sqrt(x^2 +64)*[x(1) -(x+27)(1)]/(x^2)
Set that to zero,
0 = [(x+27)/x]*[x /sqrt(x^2 +64)] +sqrt(x^2 +64)*[x(1) -(x+27)(1)]/(x^2)
0 = (x+27)/sqrt(x^2 +64) -[27sqrt(x^2 +64) /(x^2)]
Clear the fractions, multiply both sides by (x^2)*sqrt(x^2 +64),
0 = (x+27)(x^2) -27sqrt(x^2 +64)*sqrt(x^2 +64)
0 = (x+27)(x^2) -27(x^2 +64)
0 = x^3 +27x^2 -27x^2 -64
0 = x^3 -64
64 = x^3
x = 4

Hence, into (i),
B = [(x+27)/x]*sqrt(x^2 +64) --------------(i)
B = [(4+27)/4]*sqrt(4^2 +64)
B = (31/4)sqrt(80)
B = 69.318 ft, minimum ---------------answer.