Maximum volume of a cylinder with constraint

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7777ft

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Problem
A post office regulation once stated that no parcel is to be of such a size that its height plus its girth exceeds 6 feet. What is the greatest volume that can be sent by post for a package of circular cross-section?

Volume of a cylinder
[MATH]f(r,h) = \pi r^2 h[/MATH]
Constraint
[MATH] height + girth = 6\\ girth\,of\,a\,cylinder= 2 \pi r\\ h + 2 \pi r = 6 \Longrightarrow h = 6 - 2 \pi r [/MATH]
[MATH]f(r,6 - 2 \pi r) = 6 \pi r^2 - 2 \pi^2 r^3[/MATH]The function becomes a single variable derivative problem. Set [MATH]f' = 0[/MATH] yields [MATH]r=\frac{2}{\pi}[/MATH], and the volume is [MATH]\frac{8}{\pi}[/MATH] cubic feet. Easy enough.

This problem was meant to be solved as a partial derivative. How to work in the constraint and keep both independent variables?
 
I think the question was asked for a normal box which there you have 3 variable (3 length of sides) and then by the constraint you will end up with two, then with partial derivative you will reach two sets of equations which will give you value of the two variables which with the constraint you can get the third one as well
Here you only have 2 variable and with the constraint you have really one variable, I don’t think you can get to use partial derivative, or at least there is no point doing it
 
Jagulba said:
I think the question was asked for a normal box which there you have 3 variable (3 length of sides) and then by the constraint you will end up with two, then with partial derivative you will reach two sets of equations which will give you value of the two variables which with the constraint you can get the third one as well
Here you only have 2 variable and with the constraint you have really one variable, I don’t think you can get to use partial derivative, or at least there is no point doing it
Click to expand...

The full question has two parts
  1. volume of a package with a rectangular cross-section, and
  2. volume of a package with a circular cross-section.
Part (1) is a rectangular box, which I completed. Part (2) is fits the description of a cylinder. The problem comes from a chapter about partial differentiation. There must be a way of completing the problem with 2 variables.
 
Why? Maybe the point is to show the difference! How you solve a problem depends on how many variables you have.
 
See I guessed that right. what we have in rectangular box is really 3 variables then with 1 constraint which will end up with 2 variables which then partial derivative is necessary to solve it
in case of cylinder we start with 2 variables and then with 1 constraint we get to 1 variable which you cant use partial derivative
not as far as I know anyway
 
I would use Lagrange multipliers:

Objective function:

[MATH]f(C,h)=\pi\left(\frac{C}{2\pi}\right)^2h=\frac{1}{4\pi}C^2h[/MATH]
Constraint:

[MATH]g(C,h)=C+h-6=0[/MATH]
We then obtain the system:

[MATH]\frac{1}{2\pi}Ch=\lambda[/MATH]
[MATH]\frac{1}{4\pi}C^2=\lambda[/MATH]
This implies:

[MATH]\frac{1}{2\pi}Ch=\frac{1}{4\pi}C^2[/MATH]
[MATH]C(C-2h)=0[/MATH]
As we presumably want \(0<C\), we are left with:

[MATH]C=2h[/MATH]
Putting this into the constraint, we ultimately obtain:

[MATH]h=2\implies C=4[/MATH]
[MATH]f_{\max}=f(4,2)=\frac{1}{4\pi}4^2(2)=\frac{8}{\pi}[/MATH]
 
MarkFL said:
I would use Lagrange multipliers:

Objective function:

[MATH]f(C,h)=\pi\left(\frac{C}{2\pi}\right)^2h=\frac{1}{4\pi}C^2h[/MATH]
Constraint:

[MATH]g(C,h)=C+h-6=0[/MATH]
We then obtain the system:

[MATH]\frac{1}{2\pi}Ch=\lambda[/MATH]
[MATH]\frac{1}{4\pi}C^2=\lambda[/MATH]
This implies:

[MATH]\frac{1}{2\pi}Ch=\frac{1}{4\pi}C^2[/MATH]
[MATH]C(C-2h)=0[/MATH]
As we presumably want \(0<C\), we are left with:

[MATH]C=2h[/MATH]
Putting this into the constraint, we ultimately obtain:

[MATH]h=2\implies C=4[/MATH]
[MATH]f_{\max}=f(4,2)=\frac{1}{4\pi}4^2(2)=\frac{8}{\pi}[/MATH]
Click to expand...

That was very interesting, I liked the approach.

Just tried to see if I can get there so I did the partial derivative on f(r,h)= pi*r^2*h
you will then consider them as equal
pi*2*r*h=pi*r^2
r=2*h
here what I dont understand.... in your case C = 2h which means 2*pi*r= 2h and in my case it's r=2*h
where is the mistake?
 
You're using the radius rather than the circumference, and you're not taking the constraint into account.
 
Jagulba said:
That was very interesting, I liked the approach.

Just tried to see if I can get there so I did the partial derivative on f(r,h)= pi*r^2*h
you will then consider them as equal
pi*2*r*h=pi*r^2
r=2*h
here what I dont understand.... in your case C = 2h which means 2*pi*r= 2h and in my case it's r=2*h
where is the mistake?
Click to expand...

@MarkFL got all fancy using [MATH]C[/MATH]. Solution using radius for regular folk.

Object function:
[MATH]f(r,h)=\pi r^2 h[/MATH]
The constraint
[MATH] \text{height} + \text{girth} = 6\\ \text{circular girth} = 2 \pi r\\ h+2 \pi r=6\\ g(r,h)=h+2 \pi r - 6=0 [/MATH]
Lagrange
[MATH] f(r,h)+\lambda g(r,h)=0\\ \pi r^2 h + \lambda (h+2 \pi r - 6)=0 [/MATH]
Partial derivatives and system of equations
[MATH]\frac{\partial}{\partial r}[\pi r^2 h + \lambda (h+2 \pi r - 6)=0] = 2 \pi rh+2 \pi \lambda = 0[/MATH][MATH]\frac{\partial}{\partial h}[\pi r^2 h + \lambda (h+2 \pi r - 6)=0] = \pi r^2 + \lambda=0[/MATH][MATH]- \pi r^2=\lambda[/MATH][MATH]2 \pi rh+2 \pi(- \pi r^2)=0[/MATH][MATH]r(h-\pi r)=0[/MATH]
Solutions
[MATH]r=0 \text{ Discard}[/MATH][MATH]r=\frac{h}{\pi} \text{ There can be only one!}[/MATH]
Solve for height by substituting radius into constraint
[MATH] h+2 \pi \left(\frac{h}{\pi}\right)=6\\ h=2 \text{ ft} [/MATH]
Solve for radius by substituting height into the one valid solution
[MATH]r=\frac{2}{\pi} \text{ ft}[/MATH]
Maximum volume
[MATH]f\left(\frac{2}{\pi},2\right)=\pi \left(\frac{2}{\pi}\right)^2 \cdot 2=\frac{8}{\pi} \text{ft}^3[/MATH]
 
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