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anonymous

  • one year ago

The C&M Express Company is setting up some warehouses to make deliveries around the Chicago area. To fit the layout of the roads, they chop Chicagoland into identical squares each of area x . A warehouse will be built at the center of each square, and deliveries destined for a location within a square will be made from the warehouse located within the same square. The engineers tell us that to cover each square from a warehouse located at the center, the transportation costs per item are proportional to Sqrt[x] and the warehouse cost per item is proportional to 1/x . .... continued ...

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  1. anonymous
    • one year ago
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    So the overall cost of servicing a square is \[a \sqrt{x}+ \frac{b}{x}\] where a and b are positive constants. How should you set x in terms of a and b to make the overall cost of servicing a square as small as can be?

  2. anonymous
    • one year ago
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    and I guess to start, if I knew the meaning of 'set x in terms of a and b' I might have a shot at working this out... does that mean the solution will look like a= ? , b=? or I find a value for x? and any idea what a and b actually represent in real world terms?, does it even matter?

  3. dan815
    • one year ago
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    a and b are arbitrary proportionality constants

  4. dan815
    • one year ago
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    \[cost(x)=a\sqrt {x} +\frac{b}{x}\\ cost'(x)=?\]

  5. dan815
    • one year ago
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    you can set the first derivative =0 to see if there is a minimum, and check the endpoints

  6. anonymous
    • one year ago
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    awesome dan, I actually had a go at this, plotted it out and took the derivative to find the dip of the trough... is the solution then that the lowest point of the plot would be the most economical point for x ?

  7. dan815
    • one year ago
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    u can rewrite x in terms of a and b from setting the first derivative = 0

  8. dan815
    • one year ago
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    the solution is simply a function of the arbitrary constants a and b,

  9. anonymous
    • one year ago
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    damn, I have to go right now though.. being called to dinner...back in an hour.. can I hit you up then?

  10. dan815
    • one year ago
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    im about to go to sleep :)

  11. anonymous
    • one year ago
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    thank you by the way.

  12. dan815
    • one year ago
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    sure thing, someone els will come and help you though, @Luigi0210

  13. anonymous
    • one year ago
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    so if anyone is looking ... is this what you mean? \[ a -> \frac{(2*b)}{x^{\frac{3}{2}}}\] and \[ b -> \frac{(a x^{\frac{3}{2}})}{2} \]

  14. anonymous
    • one year ago
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    I took this function... \[ f[x] = \frac{b}{x} + a* \sqrt{x}\] then got the derivative as.. \[ f'[x] = -\frac{b}{x^{2}} + \frac{a}{2\sqrt{x}}​ \] then rearranged to get... \[ a -> \frac{(2*b)}{x^{\frac{3}{2}}} \] and \[ b -> \frac{(a x^{\frac{3}{2}})}{2}\]

  15. anonymous
    • one year ago
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    @Hero, @preetha hi guys, could you give this a quick look and let me know if I'm on the right track.. appreciated, thanks.

  16. anonymous
    • one year ago
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    greets @hartnn , sorry to ambush you, can I hassle you for a sec to check on my work over here?

  17. hartnn
    • one year ago
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    they are asking for x as a function of a and b x= f(a,b) which means, we need to isolate x from \(\Large f'[x] = -\dfrac{b}{x^{2}} + \dfrac{a}{2\sqrt{x}}​ = 0 \)

  18. hartnn
    • one year ago
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    "How should you set x in terms of a and b" right? so they are expecting, x = ... some terms in a and b ...

  19. hartnn
    • one year ago
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    first step \(\Large-\dfrac{b}{x^{2}} + \dfrac{a}{2\sqrt{x}}​ = 0 \\\Large \dfrac{b}{x^{2}} = \dfrac{a}{2\sqrt{x}} \) try to go ahead.. :)

  20. anonymous
    • one year ago
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    I get 3 solutions .... Here's one of them. \[ x -> \frac{ (-2)^{\frac{2}{3}}*b^{\frac{2}{3}}}{a^{\frac{2}{3}}} \]

  21. anonymous
    • one year ago
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    The other two are imaginary, so I think I can ignore them

  22. hartnn
    • one year ago
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    thats an complex number right? chose a real number from the 3 solutions

  23. hartnn
    • one year ago
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    (-2)^(2/3) is complex number

  24. hartnn
    • one year ago
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    2^(2/3) is a real number :)

  25. anonymous
    • one year ago
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    If I multiply that out.. will it be considered a real number?

  26. anonymous
    • one year ago
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    \[ x -> \frac{1.5874010519681996*b^{\frac{2}{3}}}{a^{\frac{2}{3}}} \]

  27. anonymous
    • one year ago
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    lol, I know I ask the dumbest questions

  28. hartnn
    • one year ago
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    imaginary/complex numbers come into picture when we have a negative number inside a square root ... a,b are positive, so forget them, (-2)^(2/3) does have a negative number same for \(\sqrt[3]{(-1)}\) so the only real solution is what you posted above :) though 2^(2/3) looks better, keep it that way and the only dumb questions are the questions not asked!

  29. hartnn
    • one year ago
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    and \(\Large x = (\dfrac{2b}{a})^{\dfrac{2}{3}}\) looks even better... but what looks better is subjective ;)

  30. anonymous
    • one year ago
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    ah I like that.. I had a lot of 2/3's .. was just thinking they probably factored out..

  31. anonymous
    • one year ago
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    once again.. you rocked it .. thank you

  32. hartnn
    • one year ago
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    most welcome ^_^ always happy to help :D

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