Help with more optimization please, Maximum/minimum

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Homework Help Overview

The discussion revolves around optimizing the surface area of a cylindrical can that must hold a volume of 500 cm³, with height constraints between 6 and 15 cm. Participants explore how to minimize material usage in the can's construction.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to relate the volume and surface area equations to find a minimum surface area by expressing height in terms of radius. Some participants suggest differentiating the surface area with respect to radius and setting the derivative to zero. Others discuss the implications of the height constraints on the solution.

Discussion Status

Participants are actively engaging with the problem, offering hints and guidance without providing complete solutions. There is a mix of interpretations regarding the optimization process, and some participants are questioning the treatment of constants like pi in their calculations.

Contextual Notes

There is a reminder about the forum's rules against posting complete solutions, emphasizing the importance of providing guidance instead. The discussion also touches on the need to verify whether the critical points found are consistent with the given height constraints.

rabbitstorm
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Homework Statement


A cylindrical can is to hold 500 cm^3 of apple juice the design must take into account that the height must be between 6 and 15 cm, inclusive. How should the can be constructed so that a minimum amount of material will be used in the construction? assume no waste.

Homework Equations


V is volume and SA is surface area

V= (pi)r^2h

SA=2(pi)rh + 2(pi)r^2


The Attempt at a Solution



so far this is what I have:

15>h>6

v=(pi)r^2h
500=(pi)r^2h
500/(pi)r^2=h

SA=2(pi)r(500/(pi)r^2) + 2(pi)r^2

Now I believe I need to find the derivative and set it to zero then solve for r.

The problem is that I can't quite figure out how I'm supposed to find the derivative if someone could give me a hand understanding the process I would really appreciate it as I have forgotten and my teacher is busy with the latest stuff.
Thanks,
~RS
 
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i'm only 14, but i think i can help you. i did this problem before. it's been a long time, but i remember i got the answer of ratio 1:1 for the diameter to height of the can. this is independent of the volume, which you can give the constant V and it will cancel out when you attempt to minimize the ratio.
 
Last edited:
S=2(pi)rh+2(pi)r^2. Replace h with V/((pi)r^2) so S is now a variable of just r (V is a constant) take derivative wrt to r and set to zero. solve for r. then calculate the ratio of r:h and the V will cancel out. you should get the answer 1:2. you still have to prove that the critical point gives a minimum but that is easy.

<< explicit solution deleted by berkeman >>

this is ugly, but calculate the ratio r:h and V cancels out

<< explicit solution deleted by berkeman >>

if you really want to be rigorous, calculate the second derivative of S and put r=(V/2p)^(1/3) into it and check that the result is positive, so that the extremum is indeed a minimum. that's quite messy so I'll leave you to do that.

you said that the height must be between 6 and 15 cm, inclusive. to check that r=(V/2p)^(1/3) is consistent with this restriction, you will need to replace V with 500. If it isn't, then the one of the two endpoints will give you the minimum surface area.

i find cylinders to be quite fascinating. for example, it can be twisted into a non-orientable manifold called the klein bottle. unlike the sphere and torus, it cannot be embedded in R^3, but only in R^4. I'm reading the proof of this but it is sooooooooooooo hard!
 
Last edited:
Tom1992,

Do not post complete solutions for homework problems -- it is against the PF rules that you agreed to. Our task is to be tutorial, and to provide hints and guidance that helps the student figure out the answer.
 
I think toms liking the attention, being 14 and knowing Calculus :) feels good doesn't it tom? But yea, don't post complete solutions. If you really want to know if your method/answer is correct as well, pm it to a mentor or someone smart. eg matt grime, mathwonk
 
matt grime and mathwonk are mathematicians that run this forum?
 
That do not run the forums, the Mentors and staff do. matt grime and mathwonk are mathematicians that are members of this forums, and they are very knowlegeable. They do not run it however
 
By the way I dug up some old notes and got the method was confused by the pi, this question is now solved. Have to remember that you treat pi as a variable sometimes and as a number others.
 
rabbitstorm said:
Have to remember that you treat pi as a variable sometimes and as a number others.

pi is a constant and is never treated as a variable because you don't solve for a value of pi (pi = 3.14159...) nor do you plot a function versus pi. i think what you mean is that pi often needs to be treated as a symbol that you manipulate in expressions as you do with x, and in that case yes.
 
  • #10
rabbitstorm said:
By the way I dug up some old notes and got the method was confused by the pi, this question is now solved. Have to remember that you treat pi as a variable sometimes and as a number others.
I was wondering if you meant you could not differentiate r2 and r! Glad to know that was not the case.

Tom1992 said:
pi is a constant and is never treated as a variable because you don't solve for a value of pi (pi = 3.14159...) nor do you plot a function versus pi. i think what you mean is that pi often needs to be treated as a symbol that you manipulate in expressions as you do with x, and in that case yes.
But it can mean the function [itex]\pi(x)[/itex], the number of prime numbers less than or equal to x.
 
  • #11
HallsofIvy said:
But it can mean the function [itex]\pi(x)[/itex], the number of prime numbers less than or equal to x.

true, but clearly he did not refer that any function in number theory
 
  • #12
[tex]\pi[/tex] can be used as a representative symbol for anything, let it be the number of prime numbers less than a variable, the ratio of the circumference to the diameter of a circle, or how many cakes various chefs can make. It just has to be explicitly stated in order to avoid confusion. Some uses have become standard notation, such as the Prime counting function, and the constant, but never the less there is nothing wrong with in.
 
  • #13
Tom1992 said:
true, but clearly he did not refer that any function in number theory

Are you sure you're only 14? Dang!
 

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