Optimization Part II

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Hckyplayer8

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A tin can (right circular cylinder) with top and bottom is to have volume V. What dimensions (radius of the bottom and height) give the minimum total surface area?

Well in order to find area of the bottom we will use the area of a circle which is A =(2)(pi)(r2).
The area of the "body" of the cylinder is A=(2)(pi)(r)(h)

With no other units given, I'm not sure what to do next.
 
Assume all units are the same, else there will be multiple answers. Do you see that?

So you want to minimize S= 2*pi*r2 + 2*pi*r*h

What is the constraint???
Remember V is a constant!
 
Jomo said:
Assume all units are the same, else there will be multiple answers. Do you see that?

So you want to minimize S= 2*pi*r2 + 2*pi*r*h

What is the constraint???
Remember V is a constant!
Click to expand...

I'm sorry but I don't believe I am following any of that.
 
What dimensions (radius of the bottom and height) give the minimum total surface area?
The problem outright says to minimize the surface area. Do you know what the surface area is for cylinder?
Let's start with the above.
 
Jomo said:
What dimensions (radius of the bottom and height) give the minimum total surface area?
The problem outright says to minimize the surface area. Do you know what the surface area is for cylinder?
Let's start with the above.
Click to expand...

Yes. We want to minimize (2)(pi)(r2) + (2)(pi)(r)(h).

I don't remember anything from geometry saying that the surface area of a cylinder is directly related to some fact about the cylinder...which is what we are seeking to answer in this problem, no?

It's not the radius is it? The radius should be able to be shrunk/enlarged without affecting the height of the cylinder.
 
Would you be able to do the problem if you were told that volume is 100 cubic units?

Since you need to minimise surface area, you need an expression for surface area, which you have.

But, your expression for SA has 2 variables. (see my comments in your Optimisation Part I post)

You would use the fact that V=100 to find h in terms of r (or r in terms of h) and then sub in SA formula to obtain an expression in r only (or h only).

In this problem though, you don't know what V is, but it is constant. So instead of having 100, you would just have V through your working.

Give it a go.
 
Hckyplayer8 said:
Yes. We want to minimize (2)(pi)(r2) + (2)(pi)(r)(h).

I don't remember anything from geometry saying that the surface area of a cylinder is directly related to some fact about the cylinder...which is what we are seeking to answer in this problem, no?

It's not the radius is it? The radius should be able to be shrunk/enlarged without affecting the height of the cylinder.
Click to expand...
Imagine a very tall and slim can - small radius, large height. Its volume is V. Now it starts changing - height is decreasing and radius is increasing, but the volume remains constant (V). Its surface area changes according to the function you derived. At some point during this process the function will be at its minimum. How do we find it? Just like in the other problem we use the constraint to express one variable through the other, substitute it in the function, find the derivative, etc.
 
Harry_the_cat said:
Would you be able to do the problem if you were told that volume is 100 cubic units?

Since you need to minimise surface area, you need an expression for surface area, which you have.

But, your expression for SA has 2 variables. (see my comments in your Optimisation Part I post)

You would use the fact that V=100 to find h in terms of r (or r in terms of h) and then sub in SA formula to obtain an expression in r only (or h only).

In this problem though, you don't know what V is, but it is constant. So instead of having 100, you would just have V through your working.

Give it a go.
Click to expand...

So with V = 100 I found variable height by the following

h = V / (pi)(r2) = 100 / (pi)(r2)

So SA (r) = (2)(pi)(r2) + (2)(pi)(r)(100 / (pi)(r2)
 
Hckyplayer8 said:
A tin can (right circular cylinder) with top and bottom is to have volume V. What dimensions (radius of the bottom and height) give the minimum total surface area?

Well in order to find area of the bottom we will use the area of a circle which is A =(2)(pi)(r2).
The area of the "body" of the cylinder is A=(2)(pi)(r)(h)

With no other units given, I'm not sure what to do next.
Click to expand...

Our objective function is the surface area \(S\), a function of two variables, radius \(r\) and height \(h\):

[MATH]S(h,r)=2\pi r^2+2\pi hr[/MATH]
We are given the constraint:

[MATH]V=\pi r^2h\implies h=\frac{V}{\pi r^2}[/MATH]
This will allow us to express the objective function in terms of one variable \(r\):

[MATH]S(r)=2\pi r^2+2\pi r\cdot\frac{V}{\pi r^2}=2\pi r^2+\frac{2V}{r}[/MATH]
Now we may compute the derivative (recalling \(V\) is a constant, a parameter) and equate this to zero to find the critical value(s).

[MATH]S'(r)=4\pi r-\frac{2V}{r^2}=\frac{2(2\pi r^3-V)}{r^2}=0[/MATH]
As we must have \(0<r\), what critical value for \(r\) do we obtain?
 
MarkFL said:
Our objective function is the surface area \(S\), a function of two variables, radius \(r\) and height \(h\):

[MATH]S(h,r)=2\pi r^2+2\pi hr[/MATH]
We are given the constraint:

[MATH]V=\pi r^2h\implies h=\frac{V}{\pi r^2}[/MATH]
This will allow us to express the objective function in terms of one variable \(r\):

[MATH]S(r)=2\pi r^2+2\pi r\cdot\frac{V}{\pi r^2}=2\pi r^2+\frac{2V}{r}[/MATH]
Now we may compute the derivative (recalling \(V\) is a constant, a parameter) and equate this to zero to find the critical value(s).

[MATH]S'(r)=4\pi r-\frac{2V}{r^2}=\frac{2(2\pi r^3-V)}{r^2}=0[/MATH]
As we must have \(0<r\), what critical value for \(r\) do we obtain?
Click to expand...

I'm with you through the derivative of (2)(pi)(r^2) equals (4)(pi)(r). What was done to find the derivative of 2V / r ?

At first glance I thought the quotient rule but either my math is wrong or that isn't it.
 
Hckyplayer8 said:
I'm with you through the derivative of (2)(pi)(r^2) equals (4)(pi)(r). What was done to find the derivative of 2V / r ?

At first glance I thought the quotient rule but either my math is wrong or that isn't it.
Click to expand...

I thought of that term as:

[MATH]2Vr^{-1}[/MATH]
and applied the power rule.
 
As for he rest of the problem, adding 2V to both sides and dividing 4(pi) results in r^3 = V / 2 (pi)

Which means r = 3 radical V / 2 (pi)
 
Yes, our critical value is:

[MATH]r=\sqrt[3]{\frac{V}{2\pi}}[/MATH]
Can you demonstrate that this critical value is at a minimum for the objective function?
 
MarkFL said:
Yes, our critical value is:

[MATH]r=\sqrt[3]{\frac{V}{2\pi}}[/MATH]
Can you demonstrate that this critical value is at a minimum for the objective function?
Click to expand...

Since I can't test this on a closed interval (since r is unbounded in the positive direction) then I need to use the second derivative test?
 
Hckyplayer8 said:
Since I can't test this on a closed interval (since r is unbounded in the positive direction) then I need to use the second derivative test?
Click to expand...

You could use the first derivative test simply be choosing test values to the left and the right of the critical value, but I would go with the simplicity of the second derivative test in this case.
 
MarkFL said:
You could use the first derivative test simply be choosing test values to the left and the right of the critical value, but I would go with the simplicity of the second derivative test in this case.
Click to expand...

Sorry but I have pondered this enough. How do I get the second derivative for this question? Its not the quotient rule right? So then I considered the Algebraic manipulation you performed earlier in this thread, but the (-2v) is screwing me up when I try to follow through with that process.
 
Hckyplayer8 said:
Sorry but I have pondered this enough. How do I get the second derivative for this question? Its not the quotient rule right? So then I considered the Algebraic manipulation you performed earlier in this thread, but the (-2v) is screwing me up when I try to follow through with that process.
Click to expand...

We have:

[MATH]S'(r)=4\pi r-2Vr^{-2}[/MATH]
Hence:

[MATH]S''(r)=4\pi+4Vr^{-3}[/MATH]
We can see that for all \(0<r\) we must have:

[MATH]0<S''(r)[/MATH]
This means our critical value must be at a minimum.
 
Just telling it like it is...I don't ever recall going over the whole Algebraic conversions from fractions to negative powers like what is being done this section.

Can anybody suggest a good place for me to refresh this topic?
 
This may be way too simplistic, but

[MATH]a > 0 \implies a^{-b} \equiv \dfrac{1}{a^b}.[/MATH]
Therefore,

[MATH]x > 0 \text { and } f(x) = \dfrac{1}{x^n} = x^{-n} \implies f'(x) = x^{\{(-n)-1)\}} = x^{\{-(n+1)\}} = \dfrac{1}{x^{(n+1)}}.[/MATH]
 
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