1. ## Calc (Optimization)

Heres what I did so far:

Plug 3c in for k:

$S = c/x^2 + 3c/(d-x)^2$

I differentiate, set equal to 0, and solve for x to get:

$0=-c(2x)+-3c[2(d-x)(-1)]$
$0=-2cx+6cd-3cx$
$0=-5cx+6cd$
$x=6d/5$

$x min = 6d/5 miles$

However this is not the correct answer.... Any help would be greatly appreciated. Thanks!

2. Originally Posted by pakman134

Heres what I did so far:

Plug 3c in for k:

$S = c/x^2 + 3c/(d-x)^2$

I differentiate, set equal to 0, and solve for x to get:

$0=-c(2x)+-3c[2(d-x)(-1)]$
$0=-2cx+6cd-3cx$
$0=-5cx+6cd$
$x=6d/5$

$x min = 6d/5 miles$

However this is not the correct answer.... Any help would be greatly appreciated. Thanks!
You mess up on the derivative

$S = c/x^2 + 3c/(d-x)^2=cx^{-2}+3c(d-x)^{-2}$

$\frac{dS}{dx}=-2cx^{-3}-6c(d-x)^{-3}(-1)=\frac{-2c}{x^3}+\frac{6c}{(d-x)^3}$

Good luck.

B

3. Originally Posted by TheEmptySet
You mess up on the derivative

$S = c/x^2 + 3c/(d-x)^2=cx^{-2}+3c(d-x)^{-2}$

$\frac{dS}{dx}=-2cx^{-3}-6c(d-x)^{-3}(-1)=\frac{-2c}{x^3}+\frac{6c}{(d-x)^3}$

Good luck.

B
hm.. I know this almost sounds like I'm asking you to do the problem for me... But I'm a little lost as to how you go about solving for x now.... =p
I tried playing around with it and factoring out the (d-x)^3
$0 = -2c/x^3 + 6c/(d-x)^3$
$0 = (-2c/x^3) + 6c/(d^3-3d^2x+3dx^3-x^3)$

And from here I'm lost once again... Do i set a lowest common denom now or am I completely off track...

4. ## no worries

$
\frac{dS}{dx}=-2cx^{-3}-6c(d-x)^{-3}(-1)=\frac{-2c}{x^3}+\frac{6c}{(d-x)^3}
$

multiply by LCD $x^3(d-x)^3$

$0=-2c(d-x)^3+6cx^3=-2c[(d-x)^3-3x^3]$

Note that $A^3-B^3=(A-B)(A^2+AB+B^2)$

rewriting we get...

$-2c[(d-x)^3-(\sqrt[3]{3}x)^3]$

With $A=(d-x) \mbox{ and } B=(\sqrt[3]{3}x)$

So we get...

$0=-2c[(d-x)-\sqrt[3]{3}x][(d-x)^2+(d-x)((\sqrt[3]{3}x)+((\sqrt[3]{3}x)^2]$

Setting the 1st factor equal to zero we get...

$(d-x)-\sqrt[3]{3}x$ solving for x we get

$d=x+\sqrt[3]{3}x \iff d=(1+\sqrt[3]{3})x \iff x=\frac{d}{1+\sqrt[3]{3}} \approx 0.40947d$