Help with more optimization please, Maximum/minimum

In summary: Keep up the good work :DIn summary, the problem is to find the dimensions for a cylindrical can that will hold 500 cm^3 of apple juice and use the minimum amount of material. The height of the can must be between 6 and 15 cm, inclusive. Using the equations for volume and surface area, the volume is solved for in terms of the radius and height. The surface area equation is then simplified to only include the variable of radius. Taking the derivative of the surface area with respect to the radius, setting it equal to zero, and solving for the radius gives a ratio of 1:2 for the diameter to height of the can. This is independent of the volume, which cancels out when calculating
  • #1
rabbitstorm
6
0

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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  • #2
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.
 
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  • #3
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!
 
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  • #4
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.
 
  • #5
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
 
  • #6
matt grime and mathwonk are mathematicians that run this forum?
 
  • #7
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
 
  • #8
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.
 
  • #9
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!
 

What is optimization?

Optimization is the process of finding the best solution for a problem, often involving maximizing or minimizing a certain quantity or objective function. It involves using mathematical and computational techniques to improve the efficiency and effectiveness of a system or process.

Why is optimization important?

Optimization is important because it allows us to make the most out of limited resources, save time and money, and improve the overall performance of a system or process. It is used in various fields such as engineering, economics, and computer science to solve complex problems and make informed decisions.

What is the difference between maximum and minimum optimization?

Maximum optimization involves finding the highest possible value for a given quantity or objective function, while minimum optimization involves finding the lowest possible value. Both types of optimization involve finding the optimal solution, but the direction of optimization may vary depending on the problem at hand.

What are some common techniques used for optimization?

Some common techniques used for optimization include linear and nonlinear programming, gradient descent, simulated annealing, and genetic algorithms. These techniques involve using mathematical models and algorithms to find the optimal solution for a given problem.

How can I improve my optimization skills?

To improve your optimization skills, you can practice solving different types of optimization problems, familiarize yourself with various optimization techniques, and stay updated on the latest developments in the field. It is also helpful to have a strong foundation in mathematics, programming, and problem-solving. Additionally, seeking guidance from experts and participating in optimization competitions can also help improve your skills.

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