Optimization: a sector is cut from a circle with r = 12 in.

[lucidbabble]

Hi everyone, I've been trying to do several problems for a couple of hours now, and I could really use some help! I looked through some other posts with similar concepts, but I still don't understand it. :/

1) A sector with central angle theta is cut from a circle of radius 12 inches, and the edges of the sector are brought together to form a cone. Find the magnitude of theta such that the volume of the cone is a maximum.

So far, I've know...
- that I need to find r and h values to plug into the equation for volume
r = theta/30

I am not sure how to find h. Am I supposed to use the Pythagorean Theorem? If so, am I supposed to get an obsure answer like: h = (square root of) - theta^2/900 + 144

-----
Now for another nasty one.

2) An offshore oil well is 2 kilometers off the coast. The refinery is 4 kilometers down the coast. Laying pipe in the ocean is twice as expensive as on land. What path should the pipe follow in order to minimize the cost?

First of all, what does it mean by "path"? Second... I am just not quite sure what to do.

Any help (and I mean ANY ) is much appreciated!

 

[mammothrob]

the path means how much of the pipe should be in the water and how much should be on the land. draw a line perpindicular from the oil well to the shore. now you need to find out where the pipe from the well should touch the shore. See a right triangle? Think pathagorean theorem.

make a and equation and optimize

 

[galactus]

Re: Optimization!

A sector with central angle theta is cut from a circle of radius 12 inches, and the edges of the sector are brought together to form a cone. Find the magnitude of theta such that the volume of the cone is a maximum.

So far, I've know...
- that I need to find r and h values to plug into the equation for volume
r = theta/30

I am not sure how to find h. Am I supposed to use the Pythagorean Theorem? If so, am I supposed to get an obsure answer like: h = (square root of) - theta^2/900 + 144

The volume of a cone is given by: \(\displaystyle \L\\V=\frac{1}{3}{\pi}r^{2}h\)........[1]

The radius of the circular sheet is the slant height of the cone, therefore, we can use Pythagoras:

\(\displaystyle \L\\r^{2}+h^{2}=12^{2}\)........[2]

Use the familiar \(\displaystyle s=r{\theta}\):

\(\displaystyle \L\\2{\pi}r=12(2{\pi}-{\theta})\)....[3]

Solve [2] for \(\displaystyle r^{2}\) and sub into [1]:

\(\displaystyle \L\\V(h)=\frac{1}{3}{\pi}(144h-h^{3}), \;\ 0\leq{h}\leq12\)

Differentiate:

\(\displaystyle \L\\V'(h)=\frac{1}{3}{\pi}(144-3h^{2})\)

\(\displaystyle \L\\V'(h)=0 \;\ when \;\ h=4\sqrt{3}\)

Now, you can find r. Once you have r, sub into [3] to solve for theta.

 

[soroban]

Re: Optimization!

Hello, lucidbabble!

I assume you made a sketch . . .

An offshore oil well is 2 kilometers off the coast.
The refinery is 4 kilometers down the coast.
Laying pipe in the ocean is twice as expensive as on land.
What path should the pipe follow in order to minimize the cost?

    W *
      | \
      |   \    ______
    2 |     \ √x² + 4
      |       \
      |         \
    - + - - - - - * - - - - - - * -
      A     x     P     4-x     R

The oil well is at \(\displaystyle W\), 2 km from the nearest point on shore \(\displaystyle (A).\)

The pipe will be laid underwater from \(\displaystyle W\) to point \(\displaystyle P\)
. . then along the shore from \(\displaystyle P\) to the refinery \(\displaystyle R.\)

Let \(\displaystyle x\,=\,AP\), then \(\displaystyle 4-x\,=\,PR\)

Let \(\displaystyle k\) = cost of laying 1 km of land-pipe (dollars).
Then \(\displaystyle 2k\) = cost of laying 1 km of ocean-pipe (dollars).

Using Pythagorus, the length of the ocean-pipe is: \(\displaystyle \,WP \:=\:\sqrt{x^2\,+\,4}\)
. . At \(\displaystyle 2k\) dollars/km, its cost is: \(\displaystyle \,2k\sqrt{x^2\,+\,4}\) dollars.

There will be \(\displaystyle 4\,-\,x\) km of land-pipe.
. . At \(\displaystyle k\) dollars/km, its cost is: \(\displaystyle \,k(4\,-\,x)\) dollars.


Hence, the total cost is: \(\displaystyle \,C\;=\;2k\sqrt{x^2\,+\,4}\,+\,k(4\,-\,x)\) dollars.

. . and that is the function we must minimize.

 

[lucidbabble]

Re: Optimization!

A sector with central angle theta is cut from a circle of radius 12 inches, and the edges of the sector are brought together to form a cone. Find the magnitude of theta such that the volume of the cone is a maximum.

So far, I've know...
- that I need to find r and h values to plug into the equation for volume
r = theta/30

I am not sure how to find h. Am I supposed to use the Pythagorean Theorem? If so, am I supposed to get an obsure answer like: h = (square root of) - theta^2/900 + 144

The volume of a cone is given by: \(\displaystyle \L\\V=\frac{1}{3}{\pi}r^{2}h\)........[1]

The radius of the circular sheet is the slant height of the cone, therefore, we can use Pythagoras:

\(\displaystyle \L\\r^{2}+h^{2}=12^{2}\)........[2]

Use the familiar \(\displaystyle s=r{\theta}\):

\(\displaystyle \L\\2{\pi}r=12(2{\pi}-{\theta})\)....[3]

Solve [2] for \(\displaystyle r^{2}\) and sub into [1]:

\(\displaystyle \L\\V(h)=\frac{1}{3}{\pi}(144h-h^{3}), \;\ 0\leq{h}\leq12\)

Differentiate:

\(\displaystyle \L\\V'(h)=\frac{1}{3}{\pi}(144-3h^{2})\)

\(\displaystyle \L\\V'(h)=0 \;\ when \;\ h=4\sqrt{3}\)

Now, you can find r. Once you have r, sub into [3] to solve for theta.

Thank you for your reply, galactus! I wasn't able to check this before I turned in my assignment, but I would still like clarification on how to solve it.

First of all, is this the only way to approach this problem? Another person told me that I was able to find r just by forming a proportion of theta/360 = r/12, so that is how I got r = theta/30

Also, can you explain to me the [3] step?

 

[lucidbabble]

Re: Optimization!

Hello, lucidbabble!

I assume you made a sketch . . .

An offshore oil well is 2 kilometers off the coast.
The refinery is 4 kilometers down the coast.
Laying pipe in the ocean is twice as expensive as on land.
What path should the pipe follow in order to minimize the cost?

    W *
      | \
      |   \    ______
    2 |     \ √x² + 4
      |       \
      |         \
    - + - - - - - * - - - - - - * -
      A     x     P     4-x     R

The oil well is at \(\displaystyle W\), 2 km from the nearest point on shore \(\displaystyle (A).\)

The pipe will be laid underwater from \(\displaystyle W\) to point \(\displaystyle P\)
. . then along the shore from \(\displaystyle P\) to the refinery \(\displaystyle R.\)

Let \(\displaystyle x\,=\,AP\), then \(\displaystyle 4-x\,=\,PR\)

Let \(\displaystyle k\) = cost of laying 1 km of land-pipe (dollars).
Then \(\displaystyle 2k\) = cost of laying 1 km of ocean-pipe (dollars).

Using Pythagorus, the length of the ocean-pipe is: \(\displaystyle \,WP \:=\:\sqrt{x^2\,+\,4}\)
. . At \(\displaystyle 2k\) dollars/km, its cost is: \(\displaystyle \,2k\sqrt{x^2\,+\,4}\) dollars.

There will be \(\displaystyle 4\,-\,x\) km of land-pipe.
. . At \(\displaystyle k\) dollars/km, its cost is: \(\displaystyle \,k(4\,-\,x)\) dollars.


Hence, the total cost is: \(\displaystyle \,C\;=\;2k\sqrt{x^2\,+\,4}\,+\,k(4\,-\,x)\) dollars.

. . and that is the function we must minimize.

Thanks for your reply, soroban! The image really helped because I was unable to visualize the problem before. I am still unclear on the next steps, though, so could you answer a few questions for me?

I am supposed to differentiate the function of C, right? Is this the correct derivative:
C' = k (1/(x^2 + 4)^1/2) + k

If that's correct, do I go on to find the critical numbers? Or do I something else..?

 

[galactus]

There's always more than one way to skin a related rates problem. Your method works just fine.

As for step 3, it comes from \(\displaystyle s=r{\theta}\)

You know, the length of an arc knowing the radius and sector angle.

\(\displaystyle \L\\s=2{\pi}r\) the circumference of the circle.

\(\displaystyle \L\\r{\theta}=\underbrace{12}_{\text{r}}\underbrace{(2{\pi}-{\theta})}_{\text{theta}}\)

It's a proportion. It's the same as yours.

You used degrees and I used radians.

\(\displaystyle \L\\\underbrace{2{\pi}}_{\text{360}}r=12(2{\pi}-{\theta})\)

Now, just set up your proportions on each side and you get:

\(\displaystyle \L\\\frac{{\theta}}{360}=\frac{r}{12}\)


We know the height is \(\displaystyle 4\sqrt{3}\)

This gives a radius of \(\displaystyle 4\sqrt{6}\)

Using these we find the max angle is about 1.15 radians or about 66.06 degrees