Calculus Question....

[25thhour]

Ill give you credits for solving... I keep getting zero which I know is wrong.

Here it is:

Five cubic inches of metal is melted down and made into a solid consisting of a circular cylinder of radius "r" and height "h", surmounted by a circular cone, the radius and height of which are both equal to "r". What dimensions give the minimal surface area?

this is what I got so far:

p=pi

area of cone = 1/3pr^2h
area of cylinder = pr^2h

V=5=pr^2h + 1/3pr^2h

since r=h I replaced all "r"s with "h"s

giving me V = ph^2h + 1/3ph^2h - 5

I then simplified

ph^3 + 1/3ph^3 - 5 = V

I then took D(v)/D(h)

which got me:

3ph^2 + ph^2 = V (First deriv)

simplified:

4ph^2 = V

Set first deriv to zero to find min.

= 0...

Wat do? I don't know if doing the limit as h --> 0 from the negative would work...

Edit: I may be doing shit totally wrong lol.

[leonid]

where does it say that r = h?

[25thhour]

Originally Posted by 25thhour

the radius and height of which are both equal to "r".

Right there :P

[LongGone]

You shouldn't replace all "r"s with "h"s, note that only the cone has the base radius and height the same (r). The cylinder still has height h and radius r

Ultimately, we want an equation for the total surface area of the object, i.e.

[A=pi*r^2+2pi*r*h+pi*r*s]

Where s is the slant height of the cone=(sqrt2)r

----------------------------------------------------

We want to take dA/dr (or dA/dh) of this to get the minimum value of the surface area. You took dV/dh, which is to find the min/max of the volume (which was already stated in the question to be 5)

The formula for A has r and h
We want to get rid of one of them, which we can by substituting the equation of the total volume:

[5=pi*r^2*h+(1/3)pi*r^3]

rewrite it as [h=(15-pi*r^3)/(3pi*r^2)]

now substitute h into the area formula, leaving us with only A and r in it

This (should) simplify to

[A=(1/3+sqrt2)pi*r^2+10/r]

Now, you take dA/dr, set dA/dr=0, and you then obtain your value of r required to get minimum A. Which you could then use to find h

[leonid]

Originally Posted by 25thhour

Right there :P

that's for the cone, not for the cylinder
edit: what longgone said :P

[jgano]

If I'm reading the question correctly, r=h only for the cone, NOT for the cylinder.

EDIT: double ninja'd

[25thhour]

Originally Posted by LongGone

Where s is the slant height of the cone=(sqrt2)r

Wat?

I was told to just used the formula for a right angle cone which was 1/3pir^2h?

EDIT: Nevermind, I realised that thats area not volume

I was wondering if the r = h wasn't for the whole equation.. Durp, I just misread the question.

[leonid]

I pointed out a mistake first so give all credits to me
ignore longgone

[Reincarnate]

i am prob. gonna get this shit wrong bc i am retarded

So we've got a cylinder of height h and radius r with a cone on top with radius and height r, total volume 5 cubic inches, and we want to minimize surface area.

Surface area of this cylinder = cylinder minus a circle since it's connected to the cone = SA_Cyl = pi*r^2 + 2*pi*r*h.
Surface area of the cone = cone minus the circle = pi*r*l where l^2 = r^2 + r^2 so SA_Con = pi*r*(r^2 + r^2)^.5

Vol = pi*r^2*h + (1/3)*pi*r^3 = 5
h is gay so let's get rid of that shit

pi*r^2*h = 5 - (1/3)*pi*r^3
h = (5 - (1/3)*pi*r^3) / (pi*r^2)

Sub into SA_Cyl + SA_Con = SA_Tot
SA_Tot = pi*r^2 + 2*pi*r*h + pi*r*(r^2 + r^2)^.5 =
= pi*r^2 + 2*pi*r*((5 - (1/3)*pi*r^3) / (pi*r^2)) + pi*r*(r^2 + r^2)^.5
= pi*r^2/3 + 2^.5*pi*r^2 + 10/r

derivative of that shit, set to 0
0 = 2/3*(pi+3*2^.5*pi)*r - 10/r^2
r = 0.96931239044276206
and therefore
h = 1.37081472874043897


EDIT: Verification: pi*0.96931239044276206^2*1.37081472874043897 + (1/3)*pi*0.96931239044276206^3 = 5

[emerald000]

Total volume = Vcylinder + Vcone = πr²h + πr³/3
We know that Volume = 5, so πr²h + πr³/3 = 5

Total area = Acylinder + Acone =
2πrh + πr² + πr(sqrt(r²+r²)) =
2πrh + πr² + πr(sqrt(2r²)) =
2πrh + πr² + sqrt(2)πr².

We want to minimize 2πrh + πr² + sqrt(2)πr², given that πr²h + πr³/3 = 5.

πr²h + πr³/3 = 5
πr²h = 5 - πr³/3
h = (5 - πr³/3) / (πr²)
h = (15 - πr³) / (3πr²)

Let's replace h in our area formula.

2πrh + πr² + sqrt(2)πr² =
2πr((15 - πr³) / (3πr²)) + πr² + sqrt(2)πr² =
(30 - 2πr³) / (3r) + πr² + sqrt(2)πr² =
sqrt(2)πr² + πr²/3 + 10/r

So we want to minimize y(r) = sqrt(2)πr² + πr²/3 + 10/r. The extrema of a function are found where the derivative = 0. So first, let's find the derivative.

y(r) = sqrt(2)πr² + πr²/3 + 10/r
y'(r) = 2sqrt(2)πr + 2πr/3 - 10/r²

We want to find where y'(r) = 0.

2sqrt(2)πr + 2πr/3 - 10/r² = 0
(6sqrt(2)πr³ + 2πr³ - 30) / (3r²) = 0
3sqrt(2)πr³ + πr³ -15 = 0
r³(3sqrt(2)π + π) = 15
r³ = 15 / (3sqrt(2)π + π)
r = (15 / (3sqrt(2)π + π)) ^ (1/3)

Now that we have r, we can find h.

h = (15 - πr³) / (3πr²)
h = (15 - π((15 / (3sqrt(2)π + π)) ^ (1/3))³) / (3π((15 / (3sqrt(2)π + π)) ^ (1/3))²)

This gives the decimal approximations of:
r = 0.9693 inches
h = 1.3708 inches

Might have done a mistake in the algebra, but the technique is there. If you need any help about it, I'll be glad to help.

EDIT: Woo I'm late.

EDIT2: Damn, forgot to add the bottom circle for the area of the cylinder. I will edit it in now.

EDIT3: Corrected that circle, still not the same answer as Rubix though.

EDIT4: Herp. Used cylinder height in volume of the cone...

EDIT5: Corrected. Now I have the same r as Rubix, but not the same h. I blame him.

EDIT6: Yet another mistake. Shouldn't have blamed Rubix.

[Reincarnate]

I prolly ****ed up

[LongGone]

Originally Posted by emerald000

Total volume = Vcylinder + Vcone = πr²h + πr²h/3 = 4πr²h/3

Vcone=πr^3/3, not πr²h/3 (the base radius and height of the cone are equal

[emerald000]

Corrected myself twice, but now it should be right.

[Thatskier]

So glad you guys are smmmartt.

Next year ill be coming here alot i think.... yyep ( considering anaru and rubix changed my mind to BC... im alittle more scared)

[Reincarnate]

emerald:
you go from
h = (15 - πr³) / (3πr²) up top which is right
to h = 15 / (4πr²) on the bottom which isn't
prolly why we have differing h values

If I plug my radius into your top equation I get 1.37 etc

[Reincarnate]

also emerald get your butt back into PE

[nois-or-e]

//moved to 'Homework & General Info' sub forum

[25thhour]

Originally Posted by nois-or-e

//moved to 'Homework & General Info' sub forum

I can never seem to find this darn thread... I knew it was there. Lol.

[emerald000]

Originally Posted by Reincarnate

emerald:
you go from
h = (15 - πr³) / (3πr²) up top which is right
to h = 15 / (4πr²) on the bottom which isn't
prolly why we have differing h values

If I plug my radius into your top equation I get 1.37 etc

Corrected. Forgot to update that equation after my correction.

Originally Posted by Reincarnate

also emerald get your butt back into PE

Finished school today, so I'll have a lot of free time in the next weeks to do PE and FFR coding.

[25thhour]

Well ****. I am totally effing lost with to what just happened. I am terrible with algebra so I don't know what happened when simplifying after the "h" was replaced...