anonymous
  • anonymous
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 ...
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
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?
anonymous
  • anonymous
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?
dan815
  • dan815
a and b are arbitrary proportionality constants

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dan815
  • dan815
\[cost(x)=a\sqrt {x} +\frac{b}{x}\\ cost'(x)=?\]
dan815
  • dan815
you can set the first derivative =0 to see if there is a minimum, and check the endpoints
anonymous
  • anonymous
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 ?
dan815
  • dan815
u can rewrite x in terms of a and b from setting the first derivative = 0
dan815
  • dan815
the solution is simply a function of the arbitrary constants a and b,
anonymous
  • anonymous
damn, I have to go right now though.. being called to dinner...back in an hour.. can I hit you up then?
dan815
  • dan815
im about to go to sleep :)
anonymous
  • anonymous
thank you by the way.
dan815
  • dan815
sure thing, someone els will come and help you though, @Luigi0210
anonymous
  • anonymous
so if anyone is looking ... is this what you mean? \[ a -> \frac{(2*b)}{x^{\frac{3}{2}}}\] and \[ b -> \frac{(a x^{\frac{3}{2}})}{2} \]
anonymous
  • anonymous
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 \[ b -> \frac{(a x^{\frac{3}{2}})}{2}\]
anonymous
  • anonymous
@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
  • anonymous
greets @hartnn , sorry to ambush you, can I hassle you for a sec to check on my work over here?
hartnn
  • hartnn
they are asking for x as a function of a and b x= f(a,b) which means, we need to isolate x from \(\Large f'[x] = -\dfrac{b}{x^{2}} + \dfrac{a}{2\sqrt{x}}​ = 0 \)
hartnn
  • hartnn
"How should you set x in terms of a and b" right? so they are expecting, x = ... some terms in a and b ...
hartnn
  • hartnn
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.. :)
anonymous
  • anonymous
I get 3 solutions .... Here's one of them. \[ x -> \frac{ (-2)^{\frac{2}{3}}*b^{\frac{2}{3}}}{a^{\frac{2}{3}}} \]
anonymous
  • anonymous
The other two are imaginary, so I think I can ignore them
hartnn
  • hartnn
thats an complex number right? chose a real number from the 3 solutions
hartnn
  • hartnn
(-2)^(2/3) is complex number
hartnn
  • hartnn
2^(2/3) is a real number :)
anonymous
  • anonymous
If I multiply that out.. will it be considered a real number?
anonymous
  • anonymous
\[ x -> \frac{1.5874010519681996*b^{\frac{2}{3}}}{a^{\frac{2}{3}}} \]
anonymous
  • anonymous
lol, I know I ask the dumbest questions
hartnn
  • hartnn
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!
hartnn
  • hartnn
and \(\Large x = (\dfrac{2b}{a})^{\dfrac{2}{3}}\) looks even better... but what looks better is subjective ;)
anonymous
  • anonymous
ah I like that.. I had a lot of 2/3's .. was just thinking they probably factored out..
anonymous
  • anonymous
once again.. you rocked it .. thank you
hartnn
  • hartnn
most welcome ^_^ always happy to help :D

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