A Piece of Wire

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Dec 26th 2008, 10:40 AM
Godfather

A Piece of Wire

A piece of wire 60 inches long is cut into six sections, two of one length and four of another length. Each of the two sections having the same length is bent into the form of a circle and the two circles are then joined by the four remaining sections to make a frame for a model of a right circular cylinder.
(a). Find the lengths of the sections which will make the cylinder of maximum volume.
(b). Justify (a).
Dec 26th 2008, 12:27 PM
Grandad

Cylinder with maximum volume
Hello Godfather

Quote:

Originally Posted by Godfather

A piece of wire 60 inches long is cut into six sections, two of one length and four of another length. Each of the two sections having the same length is bent into the form of a circle and the two circles are then joined by the four remaining sections to make a frame for a model of a right circular cylinder.
(a). Find the lengths of the sections which will make the cylinder of maximum volume.
(b). Justify (a).

Suppose the length of one of the two equal lengths is inches. Then:
- Find, in terms of $x$ , the radius of the circle into which one of these wires can be bent, using circumference of circle = $2\pi r$
- Find the length of one of the four equal lengths, by subtracting $2x$ from $60$ and dividing the result by $4$ . This gives the height of the cylinder in terms of $x$ .
- Use the formula for the volume of a cylinder to write down an expression for the volume, $V$ , in terms of $x$ .
- Use $\frac{dV}{dx}=0$ to find the value of $x$ which makes the value of $V$ a maximum or minimum.
- Check that this value gives a maximum.
Hope you can do it from here.

Grandad
Dec 30th 2008, 02:08 PM
OnMyWayToBeAMathProffesor

for your first point, if the length of on wire is x then the radius would by $r=\frac{x}{2\pi}$ correct?

I understand point 3,4 and 5, but not 2.

Could you please explain 2? Thanks.
Dec 30th 2008, 02:12 PM
galactus

We have 60 inches of wire. 2 pieces are of equal length and 4 of another length.

Let the 4 pieces be for the height. 60-2x is the total length of those pieces because we allowed the 2x to be for the circular top and bottom.

There are 4 pieces, so each has length

$\frac{60-2x}{4}=15-\frac{x}{2}$

See now?.
Dec 30th 2008, 02:36 PM
OnMyWayToBeAMathProffesor

thanks a lot,

So the volume if the the cylinder is $v=\pi(\frac{x}{2\pi})^2(15-\frac{x}{2})$ correct?

so i tried taking the derivative of that, using power rule, chain rule, quotient rule etc. but couldn't get an answer, any help there?
Dec 30th 2008, 05:00 PM
OnMyWayToBeAMathProffesor

so i know the derivative is

$\frac{15\pi^3x-\frac{\pi^3x^2}{2}}{2}-\frac{\pi^3x^2}{8}$

but how does one derive that. also, can that be simplified anymore?

Thanks everyone!
Dec 30th 2008, 05:24 PM
skeeter

Quote:

Originally Posted by OnMyWayToBeAMathProffesor

thanks a lot,

So the volume if the the cylinder is $v=\pi(\frac{x}{2\pi})^2(15-\frac{x}{2})$ correct?

so i tried taking the derivative of that, using power rule, chain rule, quotient rule etc. but couldn't get an answer, any help there?

$r = \frac{x}{2\pi}$

$\frac{dr}{dx} = \frac{1}{2\pi}$

$h = 15 - \frac{x}{2}$

$\frac{dh}{dx} = -\frac{1}{2}$

$V = \pi r^2 h$

$\frac{dV}{dx} = \pi \left[r^2 \cdot \frac{dh}{dx} + h \cdot 2r \cdot \frac{dr}{dx} \right]$

now substitute for r, h, and their respective derivatives and optimize.
Dec 30th 2008, 05:52 PM
OnMyWayToBeAMathProffesor

how did u get, $\frac{dV}{dx} = \pi \left[r^2 \cdot \frac{dh}{dx} + h \cdot 2r \cdot \frac{dr}{dx} \right]$ .

i thought it would have been $\frac{dV}{dx} = \pi \left[r^2 \cdot \frac{dh}{dx} + h \cdot \frac{dr}{dx} \right]$ .
Dec 30th 2008, 06:04 PM
skeeter

basic implicit derivative ...

$\frac{d}{dx} (r^2) = 2r \frac{dr}{dx}$
Dec 30th 2008, 06:14 PM
OnMyWayToBeAMathProffesor

i am so sorry, i knew that and shouldn't have missed it.

so the deriv would be $\frac{dV}{dx} = \pi \left[(\frac{x}{2\pi})^2 \cdot \frac{-1}{2} + 15-\frac{x}{2} \cdot \frac{x}{\pi} \cdot \frac{1}{2\pi} \right]$ = $\pi\left[\frac{-x^2}{4\pi}+15-\frac{x^2}{4\pi^2}\right]$ right? then i take the deriv of that? which would just be $\pi\left[\frac{-x^2}{4\pi}+15-\frac{x^2}{4\pi^2}\right]$ ?
Dec 30th 2008, 06:24 PM
skeeter

Quote:

Originally Posted by OnMyWayToBeAMathProffesor

i am so sorry, i knew that and shouldn't have missed it.

so the deriv would be $\frac{dV}{dx} = \pi \left[(\frac{x}{2\pi})^2 \cdot \frac{-1}{2} + 15-\frac{x}{2} \cdot \frac{x}{\pi} \cdot \frac{1}{2\pi} \right]$ = $\frac{-x^2}{4\pi}+15-\frac{x^2}{4\pi^2}$ right?

no ... you really need to take more care in dealing with your algebra.

$\frac{dV}{dx} = \pi \left[\left(\frac{x}{2\pi}\right)^2 \cdot \left(-\frac{1}{2}\right) + \left(15-\frac{x}{2}\right) \cdot \frac{x}{\pi} \cdot \frac{1}{2\pi} \right] = \frac{3x(20 - x)}{8\pi}$
Dec 30th 2008, 10:10 PM
Grandad

Differentiating

Hello OnMyWayToBeAMathProffesor

Quote:

Originally Posted by OnMyWayToBeAMathProffesor

thanks a lot,

So the volume if the the cylinder is $v=\pi(\frac{x}{2\pi})^2(15-\frac{x}{2})$ correct?

so i tried taking the derivative of that, using power rule, chain rule, quotient rule etc. but couldn't get an answer, any help there?

Sorry to be slow in answering. It seems that you've been going round in circles here! Keep it simple: don't use products, implicit differentiation or chain rule - just multiply out the brackets:

$v = \pi\frac{x^2}{4\pi^2}(15-\frac{x}{2})$

$= \frac{15x^2}{4\pi}-\frac{x^3}{8\pi}$

Now differentiate:

$\frac{dv}{dx}=\frac{30x}{4\pi}-\frac{3x^2}{8\pi}$

$= \frac{60x-3x^2}{8\pi}$

That was easy, wasn't it?

Then say $\frac{dv}{dx}=0$ when $60x-3x^2 = 0$

Factorise and solve for $x$ . There are two answers: one gives a maximum, and one a minimum. OK now?

Grandad
Dec 31st 2008, 09:22 AM
OnMyWayToBeAMathProffesor

Thank you so much Grandad!! and skeeter! so now I'm just going to graph it, find the zeros and check if they are a max or min.

the 2 zeros are at x=0 and x=20. x=0 is a min and x=20 is a max. thus x=20 is the correct value. Thanks again everybody!

Happy new years!