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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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
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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?
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?
a and b are arbitrary proportionality constants

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

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