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Math Help - Cylindrical can problem

  1. #1
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    Cylindrical can problem

    I am having great difficulties figuring out e), f), and g) on this.
    Could anyone help me? I am rather lost.
    Any help would be greatly appreciated.
    View image: IMG 20121126 082818
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  2. #2
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    Re: Cylindrical can problem

    (a) You are told that the can has a top and bottom consisting of circles of radius r and the can body is h high. Do you know how to find the circumference of a circle of radius r? The body will be rectangle with height h and length equal to the circumference of the circle. Do you know how to find the area of a circle of radius r and the area of a rectangle given height and length? Multiply the total area by k to find the cost. Quite frankly, that's the easy part and if you are taking a course in which you are asked to to the rest, and can't do (a), you have problems!

    (b) You are asked to show that, if the volume is given as "V", that the minimum cost is when h/r= 2 (when the height of the can is twice the radius). Do you know how to find the volume of a can or radius r and height h? If so, set that equal to V and you can solve for h as a function of r. Put that into the formula from (a) for cost and find its minimum. Do that by setting the derivative, with respect to r, equal to 0.

    (c) Instead of multiplying the cost, k, per square inch, by the actual area, multiply it by the area of the rectangular body plus twice the area of a square with sides of length r.

    (d) Do the same as (b)with this new cost function.

    (e) Add to the cost in (c) the cost of the weld, j times the length of the weld, which is two times the circumference of the circles.

    (f) Do the same as (b) with this new cost function.

    (g) Since the length of the welds is the twice circumference of the circles, if the cost of the weld per inch, j, is large compared to the cost of the metal per square inch, k, you will want to make r small and keep the same volume by make h larger. If the cost of weld per inch is small compared with the cost of metal per square inch, k, just the opposite.
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  3. #3
    Forum Admin topsquark's Avatar
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    Re: Cylindrical can problem

    Quote Originally Posted by rupert View Post
    I am having great difficulties figuring out e), f), and g) on this.
    Could anyone help me? I am rather lost.
    Any help would be greatly appreciated.
    View image: IMG 20121126 082818
    First, I know that this is all one long problem but it would be easier to look at in pieces. For example make a thread posting, say, the first three questions. Then post the next three, etc.

    Also...if you are having problems with e, f, and g then post your solutions to a, b, c, and d. That way we can see if those answers are correct (if they aren't it might explain where the problem is) as well as keeping the conversation more where it's needed.

    -Dan
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  4. #4
    MHF Contributor MarkFL's Avatar
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    Re: Cylindrical can problem

    e) The total cost is the sum of the cost of materials (including waste) and production costs (welding):

    C_T(h,r)=(2\pi rh+8r^2)k+(4\pi r+h)j

    f) subject to the constraint:

    g(h,r)=\pi r^2h-V=0

    Using Lagrange multipliers, we get the system:

    2\pi rk+j=\lambda \pi r^2

    \pi hk+8rk+2\pi j=\lambda \pi rh

    From this, you should be able to determine \frac{k}{j}.

    edit: I stepped away for a bit, and didn't realize a post had been made with good suggestions...so put this on hold for now.
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  5. #5
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    Re: Cylindrical can problem

    Thanks a lot guys. Here is my work so far.
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