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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 .
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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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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So 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?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0and 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?

dan815
 one year ago
Best ResponseYou've already chosen the best response.5a and b are arbitrary proportionality constants

dan815
 one year ago
Best ResponseYou've already chosen the best response.5\[cost(x)=a\sqrt {x} +\frac{b}{x}\\ cost'(x)=?\]

dan815
 one year ago
Best ResponseYou've already chosen the best response.5you can set the first derivative =0 to see if there is a minimum, and check the endpoints

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0awesome 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 ?

dan815
 one year ago
Best ResponseYou've already chosen the best response.5u can rewrite x in terms of a and b from setting the first derivative = 0

dan815
 one year ago
Best ResponseYou've already chosen the best response.5the solution is simply a function of the arbitrary constants a and b,

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0damn, I have to go right now though.. being called to dinner...back in an hour.. can I hit you up then?

dan815
 one year ago
Best ResponseYou've already chosen the best response.5im about to go to sleep :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0thank you by the way.

dan815
 one year ago
Best ResponseYou've already chosen the best response.5sure thing, someone els will come and help you though, @Luigi0210

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so if anyone is looking ... is this what you mean? \[ a > \frac{(2*b)}{x^{\frac{3}{2}}}\] and \[ b > \frac{(a x^{\frac{3}{2}})}{2} \]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I 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 \[ b > \frac{(a x^{\frac{3}{2}})}{2}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@Hero, @preetha hi guys, could you give this a quick look and let me know if I'm on the right track.. appreciated, thanks.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0greets @hartnn , sorry to ambush you, can I hassle you for a sec to check on my work over here?

hartnn
 one year ago
Best ResponseYou've already chosen the best response.2they are asking for x as a function of a and b x= f(a,b) which means, we need to isolate x from \(\Large f'[x] = \dfrac{b}{x^{2}} + \dfrac{a}{2\sqrt{x}} = 0 \)

hartnn
 one year ago
Best ResponseYou've already chosen the best response.2"How should you set x in terms of a and b" right? so they are expecting, x = ... some terms in a and b ...

hartnn
 one year ago
Best ResponseYou've already chosen the best response.2first 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.. :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I get 3 solutions .... Here's one of them. \[ x > \frac{ (2)^{\frac{2}{3}}*b^{\frac{2}{3}}}{a^{\frac{2}{3}}} \]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0The other two are imaginary, so I think I can ignore them

hartnn
 one year ago
Best ResponseYou've already chosen the best response.2thats an complex number right? chose a real number from the 3 solutions

hartnn
 one year ago
Best ResponseYou've already chosen the best response.2(2)^(2/3) is complex number

hartnn
 one year ago
Best ResponseYou've already chosen the best response.22^(2/3) is a real number :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0If I multiply that out.. will it be considered a real number?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[ x > \frac{1.5874010519681996*b^{\frac{2}{3}}}{a^{\frac{2}{3}}} \]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0lol, I know I ask the dumbest questions

hartnn
 one year ago
Best ResponseYou've already chosen the best response.2imaginary/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!

hartnn
 one year ago
Best ResponseYou've already chosen the best response.2and \(\Large x = (\dfrac{2b}{a})^{\dfrac{2}{3}}\) looks even better... but what looks better is subjective ;)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ah I like that.. I had a lot of 2/3's .. was just thinking they probably factored out..

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0once again.. you rocked it .. thank you

hartnn
 one year ago
Best ResponseYou've already chosen the best response.2most welcome ^_^ always happy to help :D
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