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Math Help - Optimization for a cylindrical barrel

  1. #1
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    Optimization for a cylindrical barrel

    Need help figuring this one out!

    A cylindrical barrel is to be constructed to hold 5\pi cubic feet of liquid. What are the dimensions of the barrel that minimize the amount of material used in it's construction?

    I know that  Volume = \pi * r^2 * h and SA = 2 \pi r^2 + 2 \pi r h

    So i started out trying to find h :  h = 5/r^2 (Got this from calculating  h = 5\pi / \pi r^2  from the Volume definition)

    Then I used the Area calculation : 2 \pi r^2 + 2 \pi r (5/r^2)

    Which turns out to :  A = 2\pi r^2 + 10 \pi / r

    This is where I become lost on what to do next. Any help would be appreciated!
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  2. #2
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    Re: Optimization for a cylindrical barrel

    These problems are all the same. You are trying to minimize area (since that minimizes material used). You put the function in terms of a single variable (in this case, r). Then, you take the derivative, set it equal to zero to find critical numbers, then check to see which one optimizes the problem. Also try endpoints (there are natural endpoints where 0<r and 0<h).
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  3. #3
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    Re: Optimization for a cylindrical barrel

    differentiate A with respect to r

    at the minimum/maximum, the differential will be = 0

    my final radius is the (5/2)^1/3
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    Re: Optimization for a cylindrical barrel

    Quote Originally Posted by infkelsier View Post
    differentiate A with respect to r

    at the minimum/maximum, the differential will be = 0

    my final radius is the (5/2)^1/3
    Thanks! Ok I got to the point where radius = (5/2)^1/3

    So then do I substitute that into the equation for h? Meaning h = 5/ (5/2)^1/3 ?
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  5. #5
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    Re: Optimization for a cylindrical barrel

    Quote Originally Posted by CompSci View Post
    Thanks! Ok I got to the point where radius = (5/2)^1/3

    So then do I substitute that into the equation for h? Meaning h = 5/ (5/2)^1/3 ?
    Yes, and you can simplify to h = (50)^(1/3)
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    Re: Optimization for a cylindrical barrel

    Quote Originally Posted by SlipEternal View Post
    Yes, and you can simplify to h = (50)^(1/3)
    How do you get 50^(1/3)?

    Ive been trying to figure out how to do 5 / ( (5/2)^ (1/3) )^2 since the r is squared in the h definition and r is = (5/2)^(1/3). On the test tomorrow we are not allowed to use scientific calculators (he gives us simple ones that can only add, subtract, divide, multiply etc). If you know a trick to share that got you to 50^(1/3) I'd be appreciative to know!
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  7. #7
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    Re: Optimization for a cylindrical barrel

    Oops. That was my bad, then.

    $h = \dfrac{5}{(5/2)^{2/3}} = 5^{1-2/3}2^{2/3} = 20^{1/3}$
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  8. #8
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    Re: Optimization for a cylindrical barrel

    Ok thanks, and this equals around 2.7144 so that would be the final answer then?

    The dimensions : r = 5/s ^ (1/3) X height = 20^(1/3) ? (using x to mean "by")
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  9. #9
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    Re: Optimization for a cylindrical barrel

    Yes
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  10. #10
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    Re: Optimization for a cylindrical barrel

    Quote Originally Posted by SlipEternal View Post
    Yes
    Ok, thanks again! One LAST question for the night ( i swear lol) You've been very helpful and kind tonight, I am very appreciative for your input!

    I am doing this one now :

    Suupose that  p(x) = ax^3+bx^2+cx+d for some choice of a,b,c,d. Show that p(x) has at most two relative extreme points.



    Would I have to take the derivative of p(x) and set it to 0? Even after that how would I know what the relative extreme points are?
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  11. #11
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    Re: Optimization for a cylindrical barrel

    I fell asleep. When you take the derivative, you get a quadratic equation. What do you know about the number of real solutions to a polynomial of degree 2? Hint, do a google search on the Fundamental Theorem of Algebra if you don't know the answer.
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